I'm asked to prove that$$P_{2n}(0)=(-1)^n\frac{1\cdot3\cdots(2n-1)}{2\cdot 4\cdots (2n)}$$ given that $$\frac{1}{\sqrt{1-2xu+u^2}}=\sum_{n=0}^\infty P_n(x)u^n$$
I tried this:
Let $x=0$ and use $$(u^2+1)^{-1/2}=\sum_{m=0}^\infty \begin{pmatrix} -1/2 \\ m \end{pmatrix} u^{2m}$$ $$\sum_{m=0}^\infty \begin{pmatrix} -1/2 \\ m \end{pmatrix} u^{2m}=\sum_{n=0}^\infty P_n(0)u^n$$
You have already proved the relation. Your last line implies: $$P_{2m}(0)=\binom{-1/2}{m}=\frac{(-1/2)(-3/2)\cdot \ldots\cdot(-(2m-1)/2)}{m!}=(-1)^m\frac{(2m-1)!!}{2^m\cdot m!},$$ as wanted.
As an alternative to the generating function technique, you can use the binomial theorem and the Rodrigues formula. Since: $$P_n(x)=\frac{1}{2^n\cdot n!}\frac{d^n}{dx^n}(x^2-1)^n$$ we have: $$ P_{2n}(0)=\frac{1}{4^n}[x^{2n}](x^2-1)^{2n}=\frac{(-1)^n}{4^n}\binom{2n}{n}.$$
