How is ${n \choose n/2}$ approximated in this manner? In this question, 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 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...

Using,as you did,
$$\sqrt{2 \pi n}\ \left(\frac{n}{e}\right)^n e^{\frac{1}{12n + 1}} < n! < \sqrt{2 \pi n}\ \left(\frac{n}{e}\right)^n e^{\frac{1}{12n}}$$
$$\frac{2^{n+\frac{1}{2}} e^{\frac{1}{12 n+1}-\frac{1}{3 n}}}{\sqrt{\pi n}
  }\lt \binom{n}{\frac{n}{2}} \lt \frac{2^{n+\frac{1}{2}} e^{\frac{1}{12 n}-\frac{2}{6 n+1}}}{\sqrt{\pi n} }$$
You could use
$$e^{\frac{1}{12 n+1}-\frac{1}{3 n}}=1-\frac{1}{4 n}+\frac{7}{288 n^2}-\frac{1}{3456 n^3}+O\left(\frac{1}{n^4}\right)$$
$$e^{\frac{1}{12 n}-\frac{2}{6 n+1}}=1-\frac{1}{4 n}+\frac{25}{288 n^2}-\frac{89}{3456
   n^3}+O\left(\frac{1}{n^4}\right)$$ which give very tight bounds.
For $n=10$, this gives
$$ 251.972 < 252 < 252.127 $$
Edit
I use the opportunity of the post to put an unpublised result from one of my past students, the late Dr. Bois,
$$\color{blue}{\binom{n}{\frac{n}{2}} > \frac{2^{n+\frac{1}{2}}\, e^{-\frac 1 {4n}}}{\sqrt{\pi n} }}$$ which is, for any $n$, better than the above left bound. For $n=10$, it gives $251.990$.
