I've been working on this question which asks to show that

$\displaystyle {{P}_{n}}(x)=\frac{1}{{{2}^{n}}n!}\frac{{{d}^{n}}} {d{{x}^{n}}}{{\left( {{x}^{2}}-1 \right)}^{n}}$

So first taking the n derivatives of the binomial expansions of (x2-1)n

$\displaystyle {{({{x}^{2}}-1)}^{n}}=\sum\limits_{k=0}^{n}{{{(-1)}^{k}}\frac{n!}{k!(n-k)!}{{x}^{2n-2k}}}$

$\displaystyle \frac{{{d}^{n}}}{d{{x}^{n}}}...=\sum\limits_{k=0}^ {n}{{{(-1)}^{k}}\frac{n!}{k!(n-k)!}(2n-2k)(2n-2k-1)...(2n-2k-n+1){{x}^{2n-2k}}}$
$\displaystyle =n!\sum\limits_{k=0}^{n}{{{(-1)}^{k}}\frac{(2n-2k)!}{k!(n-k)!(n-2k)!}{{x}^{2n-2k}}}$

and comparing it with

$\displaystyle {{P}_{n}}(x)=\sum\limits_{m=0}^{M}{{{(-1)}^{m}}\frac{(2n-2m)!}{{{2}^{n}}m!(n-m)!(n-2m)!}{{x}^{n-2m}}},\,\,\,\,M=\frac{n}{2},\frac{n-1}{2}$

$\displaystyle =\frac{1}{{{2}^{n}}}\sum\limits_{m=0}^{\frac{n}{2} }{{{(-1)}^{m}}\frac{(2n-2m)!}{m!(n-m)!(n-2m)!}{{x}^{n-2m}}}$

I'm having trouble with the final part,

It's clear that there's a factor of 1/n!2n difference between them but also

the Pn(x) series has m=0...n/2, and also xn , where as the n'th derivative series has k=0...n and x2n.

How can you rewrite one in terms of the other so they both have the same sum limits?

I've tried setting k=2s in the n'th derivative series and a bunch of other similar changes, but non will change the n'th powers of x.

The reason I noticed this was because the last terms of the series arn't the same,

the first series has last term, (-n)! on the bottom, means 1/infinity right?

$\displaystyle n!{{(-1)}^{n}}\frac{0!}{n!0!(-n)!}$

and the second

$\displaystyle \frac{1}{{{2}^{n}}}{{(-1)}^{\frac{n}{2}}}\frac{n!}{n!(\frac{n}{2})!0!}{{x }^{0}}$

Have I made a mistake early on or is there a clever way to combine the two series?