Product of the entries in a row of Pascal's triangle The sum of the $n$-th row in Pascal's triangle
\begin{equation}
\sum_{k=0}^{n}\binom{n}{k}
\end{equation}
has the well-known value of $2^n$. Now, I'm looking for the value of the product of the $n$-th row in Pascal's triangle, given through
\begin{equation}
s_n=\prod_{k=0}^{n}\binom{n}{k}.
\end{equation}
Any ideas how to calculate this value? Is it even possible?
I found some papers (e.g. Finding e in Pascals Triangle) dealing with the growth of this sequence, and it seems to be that the ratio of the ratios $\frac{s_{n+1}/s_n}{s_n/s_{n-1}}$ has the limiting value of 
\begin{equation}
\lim_{n\rightarrow\infty}\frac{s_{n+1}s_{n-1}}{(s_n)^2}=e.
\end{equation}
Is this helpful for calculating the value of $s_n$? So far, it is not clear to me how the growth rate of a sequence relate to its value.
 A: The closest I came to tackling this problem was through a recursive definition of the function.
Consider expanding the function
$$
\begin{equation}
s_n=\prod_{k=0}^{n}\binom{n}{k}.
\end{equation}
$$
$$
\begin{equation}
s_n=\binom{n}{0}.\binom{n}{1}.\binom{n}{2}...\binom{n}{n}
\end{equation}
$$
$$
\begin{equation}
s_n=\frac{n!}{(n-0)!(0)!}\times\frac{n!}{(n-1)!(1)!}\times\frac{n!}{(n-2)!(2)!}\times...\times\frac{n!}{(n-n)!(n)!}
\end{equation}
$$
There are $(n + 1)$ terms in this expansion. The product can also be written like this:
$$
\begin{equation}
s_n= \frac{n! \times n! \times n! \times ...\times n! }{{\Big( (n-0)! \times (n-1)! \times ... \times (1)! \times (0)!\Big) \times \Big( (0)! \times (1)! \times ... \times (n-1)! \times (n)!\Big)}}
\end{equation}
$$
$$
\begin{equation}
s_n= \frac{(n!)  ^ {n + 1}}{{( 1! \times 2! \times 3! \times... \times n!) ^ 2 }}
\end{equation}
$$
Similarly,
$$
\begin{equation}
s_{n+1}= \frac{\big((n+1)!\big)  ^ {n + 2}}{{( 1! \times 2! \times 3! \times... \times (n+1)!) ^ 2 }}
\end{equation}
$$
Simplifying the numerator and the denominator we get,
$$
\begin{equation}
\big((n+1)!\big)  ^ {n + 2} = (n+1)! \times \big( (n+1)! \big)^{n+2} = (n+1)! \times (n+1)^{n+1} \times ( n!)^{n+1}
\end{equation}
$$
And,
$$
\begin{equation}
( 1! \times 2! \times 3! \times... \times (n+1)!) ^ 2 = \big((n+1)!\big)^2 \times ( 1! \times 2! \times 3! \times... \times n!) ^ 2
\end{equation}
$$
Therefore,
$$
\begin{equation}
s_{n+1}= \frac{(n+1)! \times (n+1)^{n+1} \times ( n!)^{n+1}}{\big((n+1)!\big)^2 \times ( 1! \times 2! \times 3! \times... \times n!) ^ 2}
\end{equation}
$$
$$
\begin{equation}
s_{n+1}= \frac{(n+1)! \times (n+1)^{n+1}}{\big((n+1)!\big)^2 }
\end{equation} \times \frac{( n!)^{n+1}}{( 1! \times 2! \times 3! \times... \times n!) ^ 2}
$$
$$
\begin{equation}
s_{n+1} = \frac{(n+1)^{n+1}}{(n+1)! } \times s_n
\end{equation} 
$$
I am unsure if this could be simplified further, or if there is a relation that does not use the factorial.
