The problem i have is:  Prove that $2^n\le n!$ for all $n \in \mathbb{N},n\ge4$  Ive been trying to use different examples of similar problems like at:    First i show the base case $n=4$ is true.  Then assuming $2^k\le k!$ for some $k \in \mathbb{N},n\ge4$  For $k+1$ we have $2^{k+1}\le (k+1)!$  Rewritten as $2\cdot2^k\le k!\cdot(k+1)$  Can you not simply say $2^k\le k!$ from the inductive hypothesis, and $2\lt4\le k\lt k+1$ proving the induction step?  I am having trouble following some of what seems to me like unnecessary steps like in the example, but feel like what i did above is wrong as im of course just learning how to use induction.
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These are just the examples from my textbook, but I don't think it did not explain well.  One of the problem was to prove the inequality   $$n&lt;2^n$$  for all integers $n$.  I understand we assume if $P(n)$ is true, $P(k+1)$ is true.  Thus $k+1&lt; 2^{k+1}$  But in the text book, they added one more line that I don't understand:  $$k+1&lt;2^k+1 \le 2^k+2^k=2\cdot2^k=2^{k+1}$$  I understand  $$k+1&lt;2^{k+1}$$  But how do I know $2^k+1$ is larger than $k+1$, and smaller than $2^{k+1}$?  I have one another example just in case,  proving $2^n&lt;n!$ using induction.  $$2^{k+1}=2\cdot 2^k&lt;2\cdot k!&lt;(k+1)k!=(k+1)!$$  I understand we should start with $2^{k+1}&lt;(k+1)!$, but  Where did $2\cdot k!$ come from? Textbook says it came from inductive hypothesis, but I still don't get it.  Please advise.
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Call $P(k): k!&lt;k^k$, for $k\geq 2$ Test it out with 2, and it's true ($2&lt;4$).   Assume that $P(k)$ is true for some $k\geq2$.  Then show that $P(k+1)$ is true.   $P(k+1): (k+1)!&lt;(k+1)^{k+1}$  Ministep: $(k+1)!=k!(k+1)$ Ministep: $(k+1)^{k+1}=(k+1)^{k}(k+1)$    $k!(k+1)&lt;(k+1)^{k}(k+1)$  Can I just pull out that (k+1) and call it a day? That will make $k!&lt;(k+1)^{k}$.   We're already assuming that $k!&lt;k^k$, for some integer $k\geq 2$, and I don't think I need to prove that $k^{k}&lt;(k+1)^{k}$. I mean, I guess I could go prove that as well, but let's assume for the moment that I don't have to. Since we're assuming the original $P(k)$ is true, I should be able to write $k!&lt;k^{k}&lt;(k+1)^{k}$ right? That would prove the $k+1$ case. Is this the correct way to go about things? It seems too easy, but makes sense.  Thanks for confirmation and/or pointing out mistakes!
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I'm having difficulty solving an exercise in my course.  The question is:      Prove that $n!\geq 2^n$.   We have to do this by induction. I started like this:   The lowest natural number where the assumption is correct is $4$ as: $4! \geq 2^4 \iff 24 \ge 16$. The assumption is: $n! \ge 2^n$.   Now proof for $(n+1)$ which brings me to: $(n+1)! \ge 2^{(n+1)}$  I think I can rewrite it somehow like this:  $$ {(n+1)} \times {n!}  \stackrel{\text{(definition of factorial)}}{\ge} 2^n \times 2 $$  $$ (n+1) \times 2^n \ge 2^n \times 2 $$  Then I think I can eliminate the $2^n$ and have something like this: $n+1 \ge 2$, or $n \ge 1$.  But I think I'm wrong here some where and was hoping somebody has some advice on this. How can I prove the above assumption?  Any help would be appreciated, kind regards.
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Prove that:  $$a^{2^n} \equiv 1\pmod{2^{n+2}}$$  I have done the base case, and then have an assumption that:  $$a^{2^k} \equiv 1 \pmod {2^{k+2}}$$  I wish to prove the last step, that:  $$a^{2^{k+1}} \equiv 1 \pmod{2^{k+3}}$$  How would I do this?
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Use induction to prove the following: $1! + 2! + .... + n! &lt; (n + 1)!$  Base case:  $n = 1$  $1! &lt; 2!$ true   Inductive step: Assume that $1! + 2! + .... + k! \le  (k + 1)!$ is true   let $n = k + 1$  $1! + 2! + .... + (k + 1)! &lt; (k + 2)!$  $1! + 2! + .... + k! + (k + 1)! &lt; (k + 2)!$  $1! + 2! + .... + k! + (k + 1)! &lt; (k + 2)!$  $(k + 1)! + (k + 1)! &lt; (k + 2)!$  $2(k + 1)! &lt; (k + 2)!$  $2(k + 1)(k)(k - 1).. &lt; (k + 2)(k + 1)(k)(k - 1)....$  $2 &lt; (k + 2)$   the right is always bigger for k are non negative integer.
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So lets say I have $$ n^{2} \le n! $$  For what positive integers is this not true? $n=2$ and $ 3$  Base case? $$n=4 \implies  16 \le 24 $$  What is the inductive hypothesis and how do I show the inductive proof? Thanks
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As title says :     Prove using mathematical induction that $n &lt; 3^n$ for all positive integers $n$.   I came till here  $(k+1)&lt;3^k + 1$  From here I don't have any idea how to show $(k+1)&lt;3^{(k+1)}$ . Please help.
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I would appreciate if somebody could help me with the following problem:      For a given positive integer $n$, can we find positive integers $a$ and $k$ ($k\geq 2$) such that $2^n-1=a^k$?    The answer would be no, I  guess.  Please give me some ideas!
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I tried proving it but I'm quite skeptical about it:  $(\Rightarrow )$ Assume $x&lt;y$. Clearly, the base case for the induction is done. Assume $x^k&lt;y^k$ for some $k\in \mathbb{N}$.  Now, $x^{k+1}=x^k\cdot x&lt;y^k\cdot y=y^{k+1}$.   $(\Leftarrow )$ Assume for contradiction that $x\not&lt; y$. Then $y\le x$. Now, suppose $y\ne x$. Then $y&lt;x$. It follows that $y^n&lt;x^n$ for every $n\in \mathbb{N}$ from $(\Rightarrow )$. This contradicts our hypothesis. Also, if $y\not&lt; x$ then $x=y$. Then $y^n=x^n$ for every $n\in \mathbb{N}$ which also contradicts our hypothesis.  Is this proof correct?
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I am a CS undergrad and I'm studying for the finals in college and I saw this question in an exercise list:     Prove, using mathematical induction, that $2^n &gt; n^2$ for all integer n greater than $4$   So I started:  Base case: $n = 5$ (the problem states "$n$ greater than $4$", so let's pick the first integer that matches)  $2^5 &gt; 5^2 \implies 32 &gt; 25$ - ok!  Now, Inductive Step:  $2^{n+1} &gt; (n+1)^2$  now expanding  $2 * 2^n &gt; n^2 + 2n + 1$  Well, I can't go on, I don't know how to finish this proof. Could anyone help me with this one?  Also, I know that the statement (inductive hypothesis) is true. Simply, I'll have a number to an exponent $n$ twice, next to $n$ squared and $2n$ (which could be seen as quadratic and linear functions) which "grows" much lower than the exponential one. It could be easily "proved by example" with any $n$ greater than $4$.  I want to know how to prove it algebraically , so my professor does not freak out.  thanks in advance.
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Can you please help me and tell, how should I move on?  Can this be proved by induction?     Every natural number $n\geq 8$ can be represented as $n=3k + 5\ell$.   Thank you in advance
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How do I show via inductive proof (and without the use of the binomial theorem) that $$ {n\choose k} \in {\mathbb{N}} $$ $$\text{ for  all }\: n,k \in \mathbb{N}_0\: \text{ with }\: n≥k .$$ For my induction proof, I would show that$$  {n\choose k-1}+{n\choose k} = {n+1 \choose k}   $$ is true. However, how do I start the induction correctly with the base cases and the hypothesis? As far as I know, I cannot simply assume $$ {n\choose k-1}+{n\choose k}, $$  so how do I come to this derivation?
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I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples.  Now he showed me an exercise he was having trouble with, which states that one should prove $$n^{n+1} &gt; (n+1)^n$$  for all $n \geq 3$. I have to admit that I too have trouble showing this inequality, as in all of my attempts, my lower bound is too low. Also, I have not yet figured out, how one gets the $(n+2)^{n+1}$, primarely the number $2$ is a problem. I think this might be solved using the binomial theorem, however, I don't think they have already seen the binomial theorem in school.  Is there an easy method to show this inequality by induction not using the binomial theorem? If no: How can one show it using the binomial theorem?  Thanks for answers in advance.
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I am trying to use mathematical induction to prove  $$(2n)!\ge2^n(n!)^2\quad\text{for }n\in\mathbb{N}$$  I am stuck at the $n=k+1$ point.
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Prove by induction $\sum \frac {1}{2^n} &lt; 1$  Well supposing the base case has been shown to be true, I start with the induction step:  Suppose true for n = k:  $$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} &lt; 1$$  Want to show this is true for:  $$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} &lt; 1$$  Now this is where i am getting stuck, should i be tryig to show that  $$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} &lt; 1 +\frac{1}{2^{k+1}}$$ or should i be attempting to show that  $1 + \frac{1}{2^{k+1}} &lt; 1$ which is utter nonsense. So i am stuck on what exactly to prove here.  Note i also thought of maybe using the geometric series of $\frac {1}{2} $ as some upper bound but then that would provide me a value that was greater than 1 so it didn't work out
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I found crazy (for me at least) induction example, in fact it just would be nice to prove. (Even have problems with starting) Any hints are highly valued: $$0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $$
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My question concerns the inductive step of the proof below. I've typed out the work I've done so far. If anyone has any suggestions on how to proceed thanks in advance.  If $n\in\mathbb{N}$, then $(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^n})\ge\frac{1}{4}+\frac{1}{2^{n+1}}$.  $Proof.$ We use mathematical induction.  Base Case. For $n=1$ we see that $1-\frac{1}{2}\ge\frac{1}{4}+\frac{1}{4}$ since $\frac{1}{2}\ge\frac{1}{2}$.  Inductive Step. Let $n&gt;1$. Assume our statement is true for $n$. Now observe that  $\begin{align}(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n+1}})&amp;\ge\\ \left((1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n}})\right)(1-\frac{1}{2^{n+1}})&amp;\ge(\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$  My question is: How do we get from $\begin{align} (\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$ to $\frac{1}{4}+\frac{1}{2^{(n+1)+1}}$?  Most of what I've attempted has led to tedious calculations that do not simplify to the desired result. I've also tried making the argument that $(1-\frac{1}{2^{n+1}})$ must be $\ge$ to $\frac{1}{2}$ but then I'm stuck with $\frac{1}{8}+\frac{1}{2^{(n+1)+1}}$... Is my approach incorrect here?