Limit of exponentials Why is $n^n (n+m)^{-{\left(n+m\over 2\right)}}(n-m)^{-{\left(n-m\over 2\right)}}$ asymptotically equal to $\exp\left(-{m^2\over 2n}\right)$ as $n,m\to \infty$?

By Stirling's approximation we have $$ \binom{2n}{n+m}= \frac{(2n)!}{(n+m)!(n-m)!} \sim \frac{\sqrt{2\pi n} (2n/e)^{2n}}{\sqrt{2\pi(n+m)} \left( \frac{n+m}{e} \right)^{n+m} \sqrt{2\pi(n-m)} \left( \frac{n-m}{e} \right)^{n-m} }= \frac{1}{\sqrt{2\pi (n^2-m^2)}} \cdot \frac{(2n)^{2n} }{ (n+m)^{n+m} (n-m)^{n-m}} .$$
Now if $m$ is "small" compared to "n" then $$n^n (n+m)^{-(n+m)/2} (n-m)^{-(n-m)/2} \sim \frac{1}{2^n} \sqrt{ \sqrt{2\pi (n^2-m^2)}\binom{2n}{n+m} }.$$
We make the assumption that $m$ is small compared to $n$ precise in the sense that we take $m=\mathcal{o}(n^{2/3})$ so that we can apply the refined entropy formula found at equation 8 of the link, which gives the result.
