Evaluating $\lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}$ 
Problem statement:
If
$$ A = \lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}, $$
Find $A^2$.
Solution:1)
\begin{align*}
\prod_{k=0}^{n} \binom{n}{k}
&= \prod_{k=1}^{n} \frac{n!}{k!(n-k)!}
= \frac{(n!)^{n+1}}{(1! \cdot 2!\cdots n!)^2} \\
&= \prod_{k=1}^{n} (n+1-k)^{n+1-2k} \\
&= \prod_{k=1}^{n} \left(\frac{n+1-k}{n+1}\right)^{n+1-2k},
\end{align*}
since $\sum_{k=1}^{n} (n+1-2k) = 0$. Taking log and limit, we get
\begin{align*}
& \frac{1}{n}\sum_{k=1}^{n}\left( 1 - \frac{2k}{n+1} \right) \ln\left( 1 - \frac{k}{n+1} \right) \\
&\to \int_{0}^{1} (1 - 2x)\log(1-x) \, \mathrm{d}x
= \frac{1}{2}.
\end{align*}
Source: FITJEE AITS 2020 FT-8 Paper 1 of JEE Advanced question 53

Could anybody explain the solution?
I tried taking log both sides to bring the power down but got stuck on evaluating $$\sum_{r=0}^n \ln\binom{n}{r}$$ further I think the answer should be $e$ instead of $0.5$ which is given.

We have the limit
$$\lim_{n\to\infty}\frac{\sum_{r=0}^{n} \log \binom{n}{r}}{n(n+1)} $$
Applying Stolz Cesaro, the limit is equivalent to
$$\lim_{n\to\infty}\frac{\sum_{r=0}^{n} \log \binom{n}{r}-\sum_{r=0}^{n+1} \log \binom{n+1}{r}}{n(n+1)-(n+1)(n+2)} $$
Since $\frac{\binom{n}{r}}{\binom{n+1}{r}}=1-\frac{r}{n+1}$, the limit simplifies to
$$\lim_{n\to\infty}\frac{\sum_{r=0}^{n} \log(1-\frac{r}{n+1})}{-2(n+1)} $$
This is now a standard Riemann sum, which upon simplifying yields $\boxed{\frac12}$
