If $x_n = (\prod_{k=0}^n \binom{n}{k})^\frac{2}{n(n+1)}$ then $\lim_{n \to \infty} x_n = e$

Here is a completely different approach.

We will use $$ \lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\tag{1} $$ when the limit on the right exists.

Furthermore, $$ \begin{align} \prod_{k=0}^{n+1}\binom{n+1}{k} &=\prod_{k=0}^n\binom{n}{k}\frac{n+1}{n-k+1}\\ &=\prod_{k=0}^n\binom{n}{k}\prod_{k=1}^{n+1}\frac{n+1}{k}\tag{2} \end{align} $$ Thus, $$ \begin{align} \lim_{n\to\infty}\prod_{k=0}^n\binom{n}{k}^{\frac2{n(n+1)}} &=\lim_{n\to\infty}\left(\prod_{k=0}^n\binom{n}{k}^{\frac2n}\right)^{\frac1{n+1}}\tag{3}\\ &=\lim_{n\to\infty}\prod_{k=1}^{n+1}\left.\binom{n+1}{k}^{\frac2{n+1}}\middle/\prod_{k=1}^n\binom{n}{k}^{\frac2n}\right.\tag{4}\\ &=\lim_{n\to\infty}\left(\prod_{k=1}^{n+1}\left.\binom{n+1}{k}\middle/\prod_{k=1}^n\binom{n}{k}\right.\right)^{\frac2{n+1}}\prod_{k=0}^n\binom{n}{k}^{\frac{-2}{n(n+1)}}\tag{5}\\ &=\lim_{n\to\infty}\left(\prod_{k=1}^{n+1}\left.\binom{n+1}{k}\middle/\prod_{k=1}^n\binom{n}{k}\right.\right)^{\frac1{n+1}}\tag{6}\\ &=\lim_{n\to\infty}\left(\prod_{k=1}^{n+1}\frac{n+1}{k}\right)^{\frac1{n+1}}\tag{7}\\ &=\lim_{n\to\infty}\left.\prod_{k=1}^{n+1}\frac{n+1}{k}\middle/\prod_{k=1}^n\frac{n}{k}\right.\tag{8}\\ &=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{9}\\[12pt] &=e\tag{10} \end{align} $$ $(4)$: apply $(1)$
$(5)$: redistribute $\prod_{k=1}^n\binom{n}{k}^{-\frac2n}$
$(6)$: multiply both sides by the left side and take the square root
$(7)$: apply $(2)$
$(8)$: apply $(1)$
$(9)$: algebra

Using Riemann Sums, $$ \begin{align} &\lim_{n\to\infty}\log\left(\prod_{k=0}^n\binom{n}{k}^{\frac2{n(n+1)}}\right)\\ &=\lim_{n\to\infty}\frac2{n(n+1)}\sum_{k=0}^n\left(\sum_{j=1}^k\log(n-j+1)-\sum_{j=1}^k\log(j)\right)\tag{1}\\ &=\lim_{n\to\infty}2\sum_{k=0}^n\left(\sum_{j=1}^k\log\left(1-\frac{j}{n+1}\right)\frac1{n+1}-\sum_{j=1}^k\log\left(\frac{j}{n+1}\right)\frac1{n+1}\right)\frac1n\tag{2}\\ &=2\int_0^1\left(\int_0^x\log(1-t)\,\mathrm{d}t-\int_0^x\log(t)\,\mathrm{d}t\right)\,\mathrm{d}x\tag{3}\\ &=2\int_0^1\Big(-(1-x)\log(1-x)-x\log(x)\Big)\,\mathrm{d}x\tag{4}\\ &=-4\int_0^1x\log(x)\,\mathrm{d}x\tag{5}\\[6pt] &=1\tag{6} \end{align} $$ $(1)$: $\binom{n}{k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k(k-1)(k-2)\cdots1}$
$(2)$: distribute the $\frac1n$ and $\frac1{n+1}$; subtract $\frac{k}{n+1}\log(n+1)$ from each of the inner sums
$(3)$: $(2)$ is $\frac{n+1}{n}$ times the Riemann sum, on a $\frac1{n+1}\times\frac1{n+1}$ grid, for this double integral
$(4)$: apply $\int\log(x)\,\mathrm{d}x=x\log(x)-x+C$ twice
$(5)$: separate integral; change variables $x\mapsto1-x$; combine integrals

Therefore, $$ \lim_{n\to\infty}\prod_{k=0}^n\binom{n}{k}^{\frac2{n(n+1)}}=e\tag{7} $$