TITLE: What is a good upper bound $n^n(n-1)^{n-1}\ldots2^21^1$?
QUESTION [16 upvotes]: Given an integer $n \ge 1$, I'd like to have a not-very-loose upper bound for the integer $$u(n) := \Pi_{k=1}^n k^k =  n^n(n-1)^{(n-1)}\ldots2^21^1.$$
It's easy see that, $u(n) \le n^{n(n+1)/2}$, but this is not very interesting.
Update
We have $u(n) \le e^{\left(\frac{1}{2}n(n+1)\log\left(\frac{2n + 1}{3}\right)\right)}$, and we can't really do much better!
Indeed, using Euler-Maclaurin, we have
$ \log(u(n)) = \int_2^nx\log x dx = \frac{1}{4}n^2(2\log(n) - 1) - 2\log(2) + \frac{1}{4} + \text{error terms}$, which is comparable to the bound $\log(u(n)) \le \frac{1}{2}n(n+1)\log\left(\frac{2n + 1}{3}\right)$ in the accepted answer (see below). In particular, we can conclude that accepted answer's bound is tight!

REPLY [3 votes]: Another simple way for finding an upper bound is using the Abel's summation $$S=\sum_{k=1}^{n}k\log\left(k\right)=\frac{n\left(n+1\right)}{2}\log\left(n\right)-\frac{1}{2}\int_{1}^{n}\frac{\left\lfloor t\right\rfloor \left(\left\lfloor t\right\rfloor +1\right)}{t}dt
 $$ where $\left\lfloor t\right\rfloor 
 $ is the floor function and since $\left\lfloor t\right\rfloor >t-1
 $ we get $$S<\frac{n\left(n+1\right)}{2}\log\left(n\right)-\frac{1}{2}\int_{1}^{n}\left(t-1\right)dt
 $$ $$=\color{red}{\frac{n\left(n+1\right)}{2}\log\left(n\right)-\frac{\left(n-1\right)^{2}}{4}}.$$