The Legendre polynomials are a total basis (if normalized) of the real space $L^2[-1,1]$.
Let $w=w(t)=e^{-\frac{t^2}{2}}$, and $\frac{1}{2^nn!}\frac{d^n}{dt^n}[(t^2-1)^n]=P_n(t)\in L^2[-1,1]$.
$$\frac{1}{\sqrt{1-2tw+w^2}}=\Sigma_{n=0}^{\infty} P_n(t)w^n$$
How does one approach proving this?
Consider the formula $$f(q)=\frac{1}{\sqrt{1-q}}=(1-q)^{-\frac{1}{2}}$$
Develop this as a generalized binomial.
Use:
$$(2k+1)!!=\frac{(2k+1)!}{2^k k!}$$
Arrive at:
$$f(q)=1+\Sigma_{k=1}^{\infty}\frac{(2k-1)!}{2^{2k-1}k!(k-1)!}q^k$$
Let $q=2tw-w^2$
$$q^k=\Sigma_{j=0}^{k}\binom{k}{j}(-1)^jw^{k+j}(2t)^{k-j}$$
Collect from all $k$ the terms of $w^n$ and apply to each $k$ summand its coefficient from the $f(q)$ series to arrive at the desired result for $n$:
$$P_n(t)w^n$$
From which it follows:
$$f(2tw-w^2)=\frac{1}{\sqrt{1-2tw+w^2}}=\Sigma_{n=0}^{\infty}P_n(t)w^n$$