I am doing some self-study on measure theory and probability. This question is taken from Jacod and Protter's Essentials of Probability, Chapter 5 (Q15):
Let $X$ be a binomial random variable, $X\sim \mathrm{Binomial}(p=\frac{1}{2}, n)$, where $n=2m$. Let $a(m,k)=\dfrac{4^{m}}{2m \choose m} P(X=m+k)$. Show that $\lim_{m\to \infty} (a(m,k))^{m}=e^{-k^{2}}$.
So far I have been able to get:
\begin{align*} [a(m,k)]^{m}&=\left[ \frac{4^{m}}{2m \choose m} P(X=m+k) \right]^{m}\\ &=\left[ \frac{4^{m}}{2m \choose m} {2m \choose (m+k)}p^{m+k}(1-p)^{2m-(m+k)} \right]^{m}\\ &=\left[ 4^{m} \frac{(m!)^2}{(m-k)!(m+k)!}\left(\frac{1}{2}\right)^{m+k} \left(\frac{1}{2}\right)^{m-k} \right]^{m}\\ &=\left[\frac{(m!)^2}{(m-k)!(m+k)!}\right]^{m}\\ &\approx\left[\left(\frac{m}{\sqrt{(m-k)(m+k)}}\right)\left(\frac{m}{e}\right)^{2m}\left(\frac{e}{m-k}\right)^{m-k}\left(\frac{e}{m+k}\right)^{m+k}\right]^{m}\text{, by stirling}\\ &=\left[\frac{m^{2m+1}}{(m-k)^{m-k+0.5}(m+k)^{m+k+0.5}}\right]^{m}\\ \end{align*} However, I am unable to get the answer from here. I think I am close, but have been working with it for a while and can't get anywhere. I hate moving on without understanding where I am making a mistake (especially with self-study). If anyone feels up to the challenge any help would be greatly appreciated.