What is a good upper bound $n^n(n-1)^{n-1}\ldots2^21^1$? 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!

Using Jensen's inequality:
Letting $A= \sum_{k=1}^n k = \frac{n (n+1)}{2}$ and $B= \sum_{k=1}^n k^2 = \frac{n (n+1)(2n+1)}{6}$
We have
$$
\begin{align}
\log(u(n))
&=\sum_{k=1}^n k \log(k)  \\
&= A \sum_{k=1}^n \frac{k}{A} \log(k)  \tag{1}\\ 
&\le A \log \left(  \sum_{k=1}^n \frac{k}{A} k \right) =
A \log \left(  \frac{B}{A}  \right) \tag{2}
\end{align}
$$
Hence
$$\log(u(n)) \le \frac{n(n+1)}{2} \log\left(\frac{2 n+1}{3}\right) \tag{3}$$
The bound seems to be quite tight:

Update: as noted by comments and OP, the bound $(3)$ agrees with the true order of growth; this can be checked by applying the trapezoidal rule to the integral:
$$  -\frac{1}{4}=\int_{0}^{1} x \log(x) dx \approx \frac{1}{n+1} \sum_{k=1}^n \frac{k}{n} \log\left(\frac{k}{n}\right) $$
which gives
$$ \log(u(n))  \approx\frac{ n(n+1)}{2}\left( \log n -\frac{1}{2} \right) \tag{4}$$
If one is interested in an approximation (instead of a bound), $(4)$ might be preferable. 
Better asymptotics here (from comments).
