$$\prod_{k=0}^n (1+x^{2^k})$$ The given expression simplifies to $(1+x)(1 + x^2)...(1 + x^{2^n})$  I am not able to proceed further. How do I express this in Summation form?
My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.  Page 57. Exercise 12. Show that the following are polynomials by converting them to the form $\sum_{k=0}^{m}a_kx^k$ for a suitable $m$. In each case $n$ is a positive integer. $a)$ $(1+x)^{2n}.$ $b)$ $\frac{1-x^{n+1}}{1-x}, x\not=1.$ $c)$ $\prod_{k=0}^{n}(1+x^{2^k}).$  The attempt at a solution: a) part of the problem is pretty easy I guess, it is example of binomial theorem, so the answer would be $(1+x)^{2n}=\sum_{k=0}^{2n}(^{2n}_{k})x^k.$ Answer to part b) would be the following: $$\frac{1-x^{n+1}}{1-x}=\frac{(1-x)(1+x+x^2+\cdots+x^n)}{1-x}=1+x+x^2+\cdots+x^n=‌​\sum_{k=0}^{n}x^k.$$thanks to @DiegoMath's hint. As for part c), we have $$\prod_{k=0}^{n}(1+x^{2^k})=(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots(1+x^{2^n})$$ and I have trouble &quot;converting&quot; this to sum which would be of a form of polynomial.
Answer turned to CW, votes pre-CW era doesn't count in tag specific stat My understanding is that if an answer becomes CW (one way or another), original author gets to keep the votes pre-CW.  Does this also keep the pre-CW vote count for the tag-specific stat? It doesn't look like it, because I checked recently and suddenly my number dropped. It looks like the tag-specific stat just counts votes for non-CW answers, and doesn't "discriminate" between the pre-CW and post-CW era. If it's eventually CW, then it simply contributes 0 votes.  I'm fine with the fact that my answer became CW (in this case, through the question having 30+ answers), but I would like to keep the votes I earned before it did for the tag stat.  Otherwise I think the system is "broken". Someone can go posting 30+ answers to questions where Jon Skeet answered, and all of Jon Skeet's answers would become CW, and he'd then be 0 at all tags. 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