Finding lower/upper bounds for $\prod_{i=2}^n \log(i)$ I have a homework problem where I need to asymptotically order a set of functions, and $\prod_{i=2}^n \log(i)$ is one of them.
Is there a tight upper/lower bound for this function?
I've tried the obvious upper bound of $\prod_{i=2}^n \log(n)$, and I've tried $2^n$ as a lower bound, but it's not particularly tight..

One quick lower bound is
$$ \prod_{i=2}^n \log(i) \ge \prod_{i=n/2}^n \log(i)
\ge (\log(\tfrac n2))^{n/2} $$
which is like $c^{n\log\log n}$, matching your upper bound at this coarse level of comparison, and better than $c^n$.
