Product of binomials

Question

I am trying to compute this quantity. It seems we have to use Stirling’s formula but I don’t really succeed... Although trying to compute the quantity with a large n it seems to converge to $e^{1/2}$. Thank you for you help

Answer 1

$$\prod_{k=0}^{n}\binom{n}{k} = \frac{n!^{n+1}}{\prod_{k=0}^{n}k!^2},\qquad A_n\stackrel{\text{def}}{=}\sum_{k=0}^{n}\log\binom{n}{k}=(n+1)\log n!-2\sum_{k=0}^{n}\log k! \tag{1}$$ hence by recalling Stirling's approximation $$ \log n! = n\log n- n+\frac{1}{2}\log n +\log\sqrt{2\pi}+o(1)\tag{2}$$ we have: $$ A_{n}-A_{n-1}=n\log n -\log n! = n-\frac{1}{2}\log(n)+O(1)\tag{3}$$ since $(n^2+n)-((n-1)^2+(n-1))=2n$, by invoking Stolz theorem we have $$ \lim_{n\to +\infty}\frac{A_n-A_{n-1}}{2n}=\frac{1}{2}\quad\Longrightarrow\quad \lim_{n\to +\infty}\frac{A_n}{n^2+n}=\frac{1}{2}\tag{4}$$ then the claim is proved by simply exponentiating both sides.
You are actually dealing with the asymptotic expansion of the hyperfactorial / Barnes G-function.

As pointed out by Paramand Singh in the comments, Stirling's approximation can be completely avoided here, since Cesàro-Stolz also proves that $A_n-A_{n-1}=n+o(n)$, which is sufficient for our purposes.