I have been having some trouble solving the following problem:
Let $X$ be a Binomial $B(\frac{1}{2},n)$ random variable, where $n = 2m$. Let $$a(m, k) = \frac{4^m}{2m \choose m} \mathbb{P}(X = m + k) $$
Show that $\lim_{m \rightarrow \infty}(a(m, k))^m = e^{-k^2}$.
My attempt thus far has been the following:
$$\lim_{m \rightarrow \infty} (a(m, k))^m $$
$$= \lim_{m \rightarrow \infty} (\frac{4^m}{2m \choose m} \mathbb{P}(X = m + k))^m$$
$$= \lim_{m \rightarrow \infty} (\frac{4^m}{\frac{n!}{m!m!}} \frac{n!}{(m+k)!(m-k)!}(\frac{1}{2})^{m+k}(\frac{1}{2})^{m-k})^m$$
$= \lim_{m \rightarrow \infty} (\frac{m! m!}{(m-k)!(m+k)!})^m$$
I have tried to simplify this further to no avail.
Any help is appreciated.
EDIT following the hint (link) from the comments:
$$\lim_{m \rightarrow \infty} (\frac{m! m!}{(m-k)!(m+k)!})^m$$
$$= \lim_{m \rightarrow \infty} (\frac{(1-\frac{1}{m})(1-\frac{2}{m})...(1-\frac{k+1}{m})}{(1+\frac{k}{m})(1+\frac{k-1}{m})...(1+\frac{1}{m})})^m$$
$$= \lim_{m \rightarrow \infty} \frac{e^{-k}}{e^k}$$
$$= e^{-k^2}$$
as desired.
Where the fact that $e^x = \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n$ was used.