In [this question](https://stackoverflow.com/questions/25813440/the-asymptotic-growth-of-n-choose-floorn-2), the author asked how to get an asymptotic growth of ${n \choose n/2}$, to which the answer is as follows. First:

$$ {n \choose n/2} = \frac{n!}{(n/2)!(n- n/2)!} = \frac{n!}{(\tfrac{n}{2}!)^2} $$

Now, to find an upper bound on this, take the upper bound of [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation) for $n!$ and the lower bound for $\tfrac{n}{2}!$ which if I'm not mistaking gives the following:

$$ {n \choose n/2} \leq \frac{\sqrt{2 \pi n} (\frac{n}{e})^n e^{\tfrac{1}{12n}}}{\Big( \sqrt{2 \pi \tfrac{n}{2}} (\frac{\tfrac{n}{2}}{e})^\tfrac{n}{2} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} $$

However, the answer states this should give:

$$ {n \choose n/2} \leq \frac{e \cdot n^{n + \tfrac{1}{2}} e^{-n}}{\Big( \sqrt{2 \pi} \cdot \tfrac{n}{2}^\tfrac{n+1}{2} e^{-\tfrac{n}{2}}\Big )^2} $$

I'm confused as to how one would get from the former to the latter. Up to now I get stuck around here:

$$ {n \choose n/2} \leq \frac{\sqrt{2 \pi n} (\frac{n}{e})^n e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi \tfrac{n}{2}} (\frac{\tfrac{n}{2}}{e})^\tfrac{n}{2} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} \\
\leq \frac{\sqrt{2 \pi n} \cdot n^n e^{-n} e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi \tfrac{n}{2}} \cdot \tfrac{n}{2}^\tfrac{n}{2} e^{-\tfrac{n}{2}} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} \\
\leq \frac{\sqrt{2 \pi} \cdot n^{n + \tfrac{1}{2}} e^{-n} e^{\tfrac{1}{12n}}}
{\Big( \sqrt{2 \pi} \cdot \tfrac{n}{2}^\tfrac{n+1}{2} e^{-\tfrac{n}{2}} e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2}
$$

Now, do the latter terms just disappear due to $\lim\_{n \rightarrow \infty} \frac{e^{\tfrac{1}{12n}}}{\Big ( e^{\tfrac{1}{12\tfrac{n}{2}+1}}\Big )^2} = 0$? But in this case, why should there still be an $e$ term in the numerator? Also what happened to the $\sqrt{2 \pi}$ in the numerator? It cannot have canceled out with the one in the denominator since the latter is still there...