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 ${^n\mathrm{C}_x} = \binom{n}{x}$.

Use AM-GM inequality as a bit of trick,
$${(C_0^nC_1^nC_2^n \ldots C_{n-1}^nC_n^n)}^{\frac{1}{n+1}} \leq \frac{C_0^n+C_1^n+C_2^n +\ldots +C_{n-1}^n+C_n^n}{n+1}=\frac{2^n}{n+1}$$
$$\lim_{n\rightarrow \infty}{(C_0^nC_1^nC_2^n \ldots C_{n-1}^nC_n^n)}^{\frac{1}{n(n+1)}} \leq \lim_{n\rightarrow \infty}\frac{2}{\sqrt[n]{n+1}}= 2$$
So the only possible answer is c).
Or
$$\lim_{n\rightarrow \infty}{(C_0^nC_1^nC_2^n \ldots C_{n-1}^nC_n^n)}^{\frac{1}{n(n+1)}} \leq \lim_{n\rightarrow \infty}((2^n)^{n+1})^{\frac{1}{n(n+1)}}= 2$$
