Find
$$A = \lim_{n \to \infty} \left(\binom{n}{0} \binom{n}{1} \binom{n}{2}\cdots \binom{n}{n}\right)^{1/((n)(n+1))}$$
What I did is convert the product of $\binom{n}{k}$ into
$$\prod_{k=0}^{ k=n} \prod_{r=1}^{r=k}\frac {n-k+r}r$$ I am unable to solve it further, so please provide the solution to this problem
A first result is that the limit, provided it exists, is $\le 2$.
Indeed, one recognizes in the red part here:
$$\left(\color{red}{\left(\binom{n}{0} \binom{n}{1} \binom{n}{2}\cdots \binom{n}{n}\right)^{1/(n+1)}}\right)^{1/n}$$
the geometrical mean of the $n+1$ terms $\binom{n}{k}$ ( where $k=0,1,2...n$) that subsequently will be taken at the power $1/n$.
This geometrical mean is less than the associated arithmetical mean:
$$\frac{1}{n+1}\left(\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots \binom{n}{n}\right)=\frac{1}{n}2^n$$
Taking the power $1/n$ of this result, we get $\frac{1}{(n+1)^{1/n}}2$ whose limit is $2$.
(Thanks to @saulspatz for spotting an error of mine).
