How does one understand $0!=1$. Is it purely a theoretical value for convenience? as the definition of factorial i.e., $n!=n\times(n-1)\times(n-2)\times...1 $ actually fails here. How can it be logically understood by the definition of factorial.
54
How can I prove  from first principles that $0!$ is equal to $1$?
67803
Alright, so the idea of an exponent, $x$, is that you are multiplying its base by itself $x$ number of times. With base $5$ and $x=3$, we have that $5^3$ = $5 \cdot 5 \cdot 5$  I understand that the logarithm with base $a$ of $x = c$,  tells us that  $$a^c = x$$  and for $c =$ positive; values for $x$ are greater than $1$, and for $c =$ negative; values for $x$ are less than $1$,  and for $c = 0$, values for $x$ are...$1$.  So in short, I understand how, by means of observation of the graph of $f(x) = \log x$, we can see that $f(1) = 0$, BUT, I see no other way to understand why $x^0 = 1$, apart from the graph and everything around that very point.  I honestly cannot get my head around the idea, "$5$ times itself $0$ times... is one".  Is it that there is no fundamental answer for this but that we simply know by the graph? Or can I truly understand $x^0 = 1$ on its own?
4484
What's the smallest positive integer which is divisible by all integers $1, 2, \dots, n$?  Is there a simple way to represent the answer?  Call it $f(n)$ here.   Clearly factorial ($n!$) satisfies the condition of divisibility.  But once you get to $f(4)$ (which is $12$), factorial ($24$) is too large, because it includes as factors both $4$ and $2$, which are redundant.  Multiplying all the prime numbers up to $n$ gives you another estimate, which is this time too low, because you need to include primes multiple times when they're repeated in the factorization of the inputs.  So the answer is somewhere between factorial, and the product of primes.  Is there a simple answer?
70248
I know that   $$(k+1)! - 1 + (k+1)(k+1)! = (k+2)! - 1$$  thanks to wolframalpha, but I don't understand the steps for simplification, and I can't seem to find any rules about factorial manipulations on google.  Can someone explain this please?
98783
If any number times zero is zero and any number time infinity is infinity, then what do you get when you multiply zero times infinity? Do they cancel one another out and equal any number since any number = 0 to infinity) and any number infinity to 0? "also can 0 and infinity be one an others inverse?
79941
With $i^2=-1$, what is $i!$ ?  I'm not sure if it is $i$ or if it is an infinite product $i(i-1)(i-2)(i-3)(i-4)(i-5)\dots$  It would make sense either way.  Is it not defined?
77483
I was trying to find a proof of $a\times0 = 0$ by myself (assuming commutativity, associativity, distributivity, etc) and I came up with $$ a+0=a(1) \implies 1 = \frac{a+0}{a} = \frac aa + \frac 0a = 1 + \frac 0a \implies \frac 0a = 0 \implies a(0)=0$$ Is there anything wrong with this proof, or does it assume anything that shouldn't be assumed?
79748
I've already proved that $$\Gamma (n)= (n-1)!$$  but I don´t really know what else to do to verify that $\Gamma$ is an extension of the factorial function for real numbers (positive) Thank you! And I´m sorry for my language, I am Spanish, so thank you again for trying to understand me.
48615
So guys, how can I evaluate and prove that $$\lim_{n \to\infty} \frac{(2n-1)!!}{(2n)!!}=0.$$ Any ideas are welcomed.  $n!!$ is the double factorial, as explained in this   post.
12098
From its definition a combination $\binom{n}{k}$, is the number of distinct subsets of size $k$ from a set of $n$ elements.  This is clearly an integer, however I was curious as to why the expression  $$\frac{n!}{k!(n-k)!}$$ always evaluates to an integer.  So far I figured:  $n!$, is clearly divisible by $k!$, and $(n-k)!$, individually, but I could not seem to make the jump to proof that that $n!$ is divisible by their product.
756
As a physics student, I have occasionally run across the gamma function  $$\Gamma(n) \equiv \int_0^{\infty}t^{n-1}e^{-t} \textrm{d}t = (n-1)!$$  when we want to generalize the concept of a factorial.  Why not define the gamma function so that   $$\Gamma(n) = n!$$  instead?  I realize either definition is equally good, but if someone were going to ask me to choose one, I would choose the second option.  Are there some areas of mathematics where the accepted definition looks more natural?  Are there some formulas that work out more cleanly with the accepted definition?
112786
I've started studying calculus.  As part of studies I've encountered a question. How does one prove that $1 &gt; 0$?  I tried proving it by contradiction by saying that $1 &lt; 0$, but I can't seem to contradict this hypothesis.  Any help will be welcomed.
113604
I was messing around with my TI-84 Plus Silver Edition calculator and discovered that it will actually give me values when taking the factorial of any number $n/2$ where $n$ is any integer greater than $-2$. Why does this happen? I thought factorials were only defined for positive integers and $0$, so what is my calculator doing to get the answer $3.32335097$ when I enter $2.5!$? Is there actually a definition of $2.5!$ or is my calculator just being weird? How is the factorial function implemented?  I understand the binary implications of $2.5$, so that could possibly have something to do with it. I get a domain error when trying to take the factorial of $-1, 2.3, e$, and any number that is not of the form $n/2$ where $n$ is any integer greater than $-2$.
76183
Okay, I'm not much of a mathematician (I'm an 8th grader in Algebra I), but I have a question about something that's been bugging me.  I know that $0.999 \cdots$ (repeating) = $1$. So wouldn't $1 - \frac{1}{\infty} = 1$ as well? Because $\frac{1}{\infty} $ would be infinitely close to $0$, perhaps as $1^{-\infty}$?  So $1 - 1^{-\infty}$, or $\frac{1}{\infty}$ would be equivalent to $0.999 \cdots$? Or am I missing something? Is infinity something that can even be used in this sort of mathematics?
14417
Why don't we define $\frac 10$ as $j$ , $\frac 20$ as $2j$ , and so on? I know that by following the rules of math this eventually leads to $1=2$ , but we could make an exception and say that $j$ is the only number such that $0*j \not= 0$ , and put other restrictions necesary so that we don't get contradictions. We do this for $i$ , so why can't we do it here? For example, $i^2$ , is defined as $-1$ , but you could also say $i^2=\sqrt {-1} *\sqrt {-1}=\sqrt {(-1)(-1)}=\sqrt{1}=1$ , but we make an exception for this.
35445
I have a feeling this question is a duplicate, but it's not coming up in "Questions that may already have your answer."  We all know very well that $F(0) = 0$, $F(1) = 1$ and $F(n) = F(n - 2) + F(n - 1)$ for all $n &gt; 1$.  I'm wondering about $n &lt; 0$. My first thought was $F(-n) = -F(n)$, which is appealing from a multiplicative point of view, as it seems to preserve certain identities, like $F(n)^2 = F(n - 1) F(n + 1) - (-1)^n$.  But it doesn't quite make sense from an additive point of view, it doesn't seem to work both "forwards" and "backwards." For example, it would give us $F(-1) + F(0) = -1 \neq F(1)$.  How do we extend $F(n)$ to negative $n$ so as to maintain both the related identities and the basic defining identity?
11249
I have an equation containing factorials  here  $$(k+1)!+(k+1)(k+1)!$$  yet I am having a hard time understanding how to simplify it using algebra. A simple search on wolfram gets me a reduced form of   $$(k + 2)! $$  This would be a great refresher to such problems, sadly I don't know the elementary operations to reduce it.