Closed form of $\prod_{i=0}^{N}\big(i!\big)^{{N}\choose{i}}$

Question

I have made a question here about closed form of the following: $$\prod_{k=0}^{N}\big(k!\big)^{{N}\choose{k}}$$ I know that there is a known closed form for, $$\prod_{i=0}^{N}\big(i!\big)=G(N+2)$$ where we have expressed it as Barnes G-function. But I can not figure out the closed form for $\prod_{k=0}^{N}\big(k!\big)^{{N}\choose{k}}$. So far I went about to write the product as Gamma functions such that, $$\prod_{k=0}^{N}\big(k!\big)^{{N}\choose{k}}=\prod_{k=0}^{N}\big(\Gamma[k+1]\big)^{\frac{\Gamma[N+1]}{\Gamma[k+1]\Gamma[N-k+1]}}$$ but this has not been helpful than being rather a redefinition. I wonder if any one has an solution to this?

Answer 1

Since this touches on a problem of personal interest (and in the hopes of encouraging a more informed answer), I will post some hand waving which suggests a negative answer: there is no elementary closed form. Of course, there may be some other function like the Barnes $G$ function mentioned in the question which allows a curt algebraic expression, but this would be a nonelementary closed form in my view.

Let us define $S(N,j) = \sum_{0 \leq k \leq j} \binom{N}{k}$, the sum of the first $j+1$ (or last) binomial coefficients in the row for $N$ of Pascal's triangle. $S(N,j)$ is known (to Knuth et. al., see Concrete Mathematics and Gosper's algorithm) to not have an elementary closed form, and there are at least four questions on MathOverflow on how to approximate $S(N,j)$. I am willing to bet \$5 USD that I will not see an elementary and correct closed form for $S(N,j)$ in my lifetime. (How I would personally collect on such a bet is another unsolved problem.)

Let us break the poster's product into terms, using the organization given by factorials. I use $M=2^N$. There are $M$ terms for ($0!=1$), $M-1$ terms which are 1, $M-1-N$ terms which are 2, and in general $M - S(N,j-1)$ terms which are $j$. So a closed form for the given product would imply a closed form for another product with terms like $j^{S(N,j-1)}$.

Now this closed form would give a new relationship between $S(N,j)$ for fixed $N$ and many $j$. It is very unclear to me that this would improve upon our knowledge of $S(N,j)$ to the point of finding a better expression for it. Thus I make a side bet of \$1 USD that I won't see (a correct derivation of) an elementary closed form for the product in the above question, either this year or in all of the year 2020 either.

Gerhard "Prefers Collecting On His Bets" Paseman, 2019.12.20.

Answer 2

First, let's consider prime factorization: $$\prod_{k=0}^N k!^\binom{N}{k} = \prod_p p^{\sum_{k=0}^N \nu_p(k!)\binom{N}{k}},$$ where the product is taken over the primes $p$.

Recalling that $$\nu_p(k!) = \left\lfloor \frac{k}{p} \right\rfloor + \left\lfloor \frac{k}{p^2} \right\rfloor + \dots,$$ and noticing that for any $q$, $\left\lfloor \frac{k}{q} \right\rfloor$ equals the coefficient of $x^k$ in $\frac{x^q}{(1-x)(1-x^q)}$, we conclude that $\sum_{k=0}^N \nu_p(k!)\binom{N}{k}$ equals the coefficient of $x^N$ is $$\sum_{j\geq 1} (1+x)^N\frac{x^{p^j}}{(1-x)(1-x^{p^j})}.$$ Using Lagrange inversion, we get that this is also the coefficient of $x^N$ in $$\sum_{j\geq 1} \frac{x^{p^j}}{(1-2x)((1-x)^{p^j}-x^{p^j})}.$$ Hence, we got the generating function for the powers of each prime $p$ in the expression of interest.