I'm working on some math in my spare time and wanted to see if it were possible to find $\prod_{n=1}^{m}(n!)^n$ as a closed form expression.
I was able to work through it to get it down to $$\prod_{n=1}^{m}(n!)^n = \prod_{n=1}^{m}(n)^{1/2(m^2+m+n-n^2)}.$$
This can go further to $$(m!^{(m+m^2)} H(m)\prod_{n=1}^{m}n^{-n^2})^{1/2},$$ where $H(m)$ is the hyperfactorial of $m$.
But I'm still unable to figure out a closed form of $\prod_{n=1}^{m}n^{-n^2}$. Any help would be appreciated, thank you.