<s>[INST] If $x_n = (\prod_{k=0}^n \binom{n}{k})^\frac{2}{n(n+1)}$ then $\lim_{n \to \infty} x_n = e$ I try to prove the following:
$$x_n = \left(\prod_{k=0}^n \binom{n}{k}\right)^\frac{2}{n(n+1)}$$
$$\lim_{n \to \infty} x_n = e$$
I want to use double sided theorem, so I've proven that
$$x_n \ge \frac{n}{\sqrt[n]{n!}}$$
As it known, $\lim\limits_{n \to \infty} \frac{n}{\sqrt[n]{n!}} = e$. But I can't find such sequence $y_n$ that $x_n \le y_n$ and $\lim\limits_{n \to \infty} y_n = e$.
There is proof for lower bound:
$$x_n = \frac{(n!)^\frac{2}{n}}{(0!\cdot 1! \dots n!)^\frac{4}{n(n+1)}} \ge \frac{(n!)^\frac{2}{n}\cdot n}{(0!\cdot 1! \dots n!)^\frac{4}{n(n+1)}\cdot(n^n(n-1)^{n-1}(n-2)^{n-2}\dots2^2)^\frac{4}{n(n+1)}} = $$
$$= \frac{(n!)^\frac 2n \cdot n}{(n!)^\frac 4n} = \frac{n}{(n!)^\frac 2n} \ge \frac{n}{\sqrt[n]{n!}}$$
 [/INST] Too late, more than likely too advanced but done for the fun of deriving asymptotics.
$$\prod_{k=0}^n \binom{n}{k}=\frac{\big[\Gamma (n+1)\big]^n}{G(n+1)\, G(n+2)}$$ where $G(.)$ is  the Barnes G-function.
Using Stirling-like approximations
$$\log(G(p))=\frac{1}{4} p^2 (2 \log (p)-3)+\frac{1}{2} p (-2 \log (p)+2+\log (\pi )+\log (2))+$$ $$\frac{1}{12} \left(-6 \log \left(2 \pi  A^2\right)+5 \log
   (p)+1\right)-\frac{1}{12 p}-\frac{1}{240 p^2}+O\left(\frac{1}{p^3}\right)$$
$$\log\Bigg[\prod_{k=0}^n \binom{n}{k} \Bigg]=\frac{n^2}{2}+\frac{1}{2} n (-\log (n)+2-\log (\pi )-\log
   (2))+\frac{1}{12} (24 \log (A)-4 \log (n)-1-6 \log (2 \pi
   ))-\frac{1}{12 n}+\frac{1}{180 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log(x_n)=1-\frac 1 n \log \left(\frac{2 \pi  n}{e}\right)+\frac{24 \log (A)+2 \log (n)-7}{6 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\color{blue}{x_n=e\left(1-\frac 1 n \log \left(\frac{2 \pi  n}{e}\right) \right)+O\left(\frac{1}{n^2}\right)}$$
The relative error is less than $0.1$% as soon as $n>135$.
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