<s>[INST] Asymptotics of ${2^n \choose n}$? How can one compute the asymptotics of ${2^n \choose n}$?  I know it is bounded below and above by $\left(\frac{2^{n}}{n}\right)^n$  and $\left(\frac{2^{n}e}{n}\right)^n$.
If I plug in Stirling's approximation I get
$$\frac{2^{2^n n+n/2-1/2}}{(2^n-n)^{2^n-n+1/2} n^{n+1/2}\sqrt{\pi}}.$$
I am hoping there is a simpler asymptotic formulation and in particular I would like to compare it to $2^{n^2}$.
 [/INST] Writing a product as the exponential of the sum of the logarithms is often a fruitful method.
Here we can write
$$\begin{align}
\binom{2^n}{n} &= \prod_{m=1}^n \frac{2^n - (m-1)}{m}\\
&= \frac{2^{n^2}}{n!} \prod_{k=1}^{n-1} \left(1- \frac{k}{2^n}\right)\\
&= \frac{2^{n^2}}{n!} \exp \left(\sum_{k=1}^{n-1} \log \left(1-\frac{k}{2^n} \right)\right)\\
&= \frac{2^{n^2}}{n!} \exp \left(-\frac{n(n-1)}{2^{n+1}} + O\left(\frac{n^3}{2^{2n}}\right)\right)
\end{align}$$
to obtain an expression that allows good bounds. We have the first result
$$\binom{2^n}{n}\sim \frac{2^{n^2}}{n!}$$
by recognising that $\exp \left(- \frac{n^2}{2^n}\right)$ can be reasonably approximated by $1$, and using Stirling's approximation for the factorial, we can write that as
$$\binom{2^n}{n} \sim \frac{1}{\sqrt{2\pi n}}\left(\frac{2^ne}{n}\right)^n.$$
Using some terms of the approximation of the logarithms, more precise expressions can be obtained, e.g.
$$\binom{2^n}{n} \approx \frac{2^{n^2}}{n!} \left(1 - \frac{n(n-1)}{2^{n+1}}\right)$$
by approximating $\log (1-x) \approx -x$ for small $x$.
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