Evaluating $\lim_{n \to \infty}\left({^n\mathrm{C}_0}{^n\mathrm{C}_1}\dots{^n\mathrm{C}_n}\right)^{\frac{1}{n(n+1)}}$ 
$\lim_{n \to \infty}\left({^n\mathrm{C}_0}{^n\mathrm{C}_1}\dots{^n\mathrm{C}_n}\right)^{\frac{1}{n(n+1)}}$ is equal to:
a) $e$
b) $2e$
c) $\sqrt e$
d) $e^2$

Though it looks really innocent at first sight, it certainly isn't.
Attempt: It's $\infty^{\infty}$ form.
I had tried taking the product raised to the power $\frac{1}{n(n+1)}$ as function $f(n)$. Then I took logarithm of both sides to see if things simplifying.  Even after factoring out the extra factorials it wasn't easy.
Note that ${^n\mathrm{C}_x} = \binom{n}{x}$.

$$\prod_{k=0}^{n}\binom{n}{k}=\frac{n!^{n}}{\prod_{k=0}^{n}k!^2}=\frac{n!^n}{\left[\prod_{k=1}^{n}k^{n+1-k}\right]^2}=\frac{n!^n}{n!^{2n+2}}\prod_{k=1}^{n}k^{2k} \tag{1}$$
hence the outcome depends on the asymptotic behaviour of the hyperfactorial.Since by Riemann sums
$$ \lim_{n\to +\infty} \frac{1}{n}\sum_{k=1}^{n}\log\frac{k}{n}=\int_{0}^{1}\log(x)\,dx = -1, $$
$$ \lim_{n\to +\infty} \frac{1}{n}\sum_{k=1}^{n}\frac{k}{n}\log\frac{k}{n}=\int_{0}^{1}x\log(x)\,dx = -\frac{1}{4}\tag{2} $$
we have
$$ \frac{1}{n^2}\log\prod_{k=0}^{n}\binom{n}{k} = \frac{1}{n^2}\left[2\sum_{k=1}^{n}k\log k-(n+2)\log n!\right]\to \frac{1}{2}\tag{3} $$
and the correct option is c).
