Product of the entries in a row of Pascal's triangle

Question

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.

Answer 1

This is OEIS A001142 which begins $$1, 1, 2, 9, 96, 2500, 162000, 26471025, 11014635520, 11759522374656, 32406091200000000, 231627686043080250000, 4311500661703860387840000, 209706417310526095716965894400, 26729809777664965932590782608648192$$ An approximate formula is given $a(n) \approx A^2 * \exp(n^2/2 + n - 1/12) / (n^{(n/2 + 1/3)} * (2*\pi)^{((n+1)/2))}$, where $A = A074962 = 1.2824271291\ldots$ is the Glaisher-Kinkelin constant.

The growth rate is dominated by the term $$\frac{\exp(\frac {n^2}2)}{n^{\frac n2}}=\exp \left( \frac{n^2}2-\frac n2\log n \right)$$

Answer 2

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.