Legendre polynomial to show identity, can't spot mistake Using Legendre polynomial generating function
\begin{equation}
\sum_{n=0}^\infty P_n (x) t^n  = \frac{1}{\sqrt{(1-2xt+t^2)}}
\end{equation}
Or $$ P_n(x)=\frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2-1)^n]  $$    
Show$$ P_{2n}(0)=\frac{(-1)^n (2n)!}{(4)^n (n!)^2}     $$ 
And $$ P_{2n+1}(0)=0$$
I expressed $$(x^2 -1)^{2n}= \sum_{k=0}^{2n} {2n \choose k}x^{4n-2k}(-1)^k$$ 
And using second formula given, the only term that remains after differentiating 2n times and substituting x=0 is where $$4n-2k=2n$$ so $$2n=2k$$ k=n i.e $$(-1)^n \frac{(2n)!}{(n!)^2} $$ but multiplying this by$$ \frac{1}{2^{2n} (2n)!} $$ doesn't give desired solution. Where have I gone wrong?

\begin{equation}
\begin{split}
P_n(x) &= \frac{1}{2^nn!}[(x+1)^n\cdot(x-1)^n]^{(n)}
\\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}\binom{n}{k}[(x+1)^n]^{(k)} \cdot [(x-1)^n]^{(n-k)}
\\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}\binom{n}{k}\frac{n!}{(n-k)!}(x+1)^{n-k} \cdot \frac{n!}{k!}(x-1)^k
\\ &= \frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}^2(x+1)^{n-k}(x-1)^k
\end{split}
\end{equation}
so 
\begin{equation}
\begin{split}
P_{2n}(0) &=\frac{1}{2^{2n}}\sum_{k=0}^{2n}\binom{2n}{k}^2(-1)^k = \frac{1}{2^{2n}}\binom{2n}{n}(-1)^n = \frac{(-1)^n(2n)!}{2^{2n}n!^2}
\end{split}
\end{equation}
and
\begin{equation}
\begin{split}
P_{2n+1}(0) &=\frac{1}{2^{2n+1}}\sum_{k=0}^{2n+1}\binom{2n+1}{k}^2(-1)^k = \frac{1}{2^{2n}}\cdot 0 = 0
\end{split}
\end{equation}
See legendre polynomial problem (pls help!) for the derivation of the value of $\sum_{k=0}^m\binom{m}{k}^2(-1)^k$.
