I'm looking for the most simple form, perhaps without product symbol, of $\displaystyle \prod_{k=1}^{n}k^k$ for any positive number $n$. Maybe this is already the most simple form?
What I've done so far is:
$$\displaystyle \prod_{k=1}^{n}k^k =\prod_{k=1}^{n}e^{k \ln (k)} = e^{\sum_{k=1}^n k\ln(k)}$$
And from there I've search online, for instance [here], an easier expression of $\displaystyle \sum_{k=1}^n k \ln(k)$ but couldn't manage to find one.
We know that $\displaystyle \sum_{k=1}^n \ln(k) = \ln(n!)~$ but we here need $k\ln(k)$.
So my guess is either I didn't search well, either there is no more simple expression of $\displaystyle \sum_{k=1}^n k \ln(k)$ depending of $n$ and without $\sum$ for instance, or the solution isn't going trough exponential form. Could you help me find out?
Since $$ \prod_{k=0}^{n-1}\frac{n!}{k!} = \frac{n!^n}{G(n+1)} $$ where $G(n)$ is the Barnes' G function, it is easy to extend the original definition to non-integer $n.$ Furthermore, the asymptotics of both the $\Gamma$ and $G$ functions are known, so the OP's expression has an easily derived asymptotic formula. For even moderate $n,$ say,$ n>6,$ this is more than likely preferred.