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..

By partial summation $$\sum_{i=2}^{n}\log\left(\log\left(i\right)\right)=n\log\left(\log\left(n\right)\right)-\int_{2}^{n}\frac{\left\lfloor t\right\rfloor }{t\log\left(t\right)}dt
 $$ where $\left\lfloor t\right\rfloor =t-\left\{ t\right\} 
 $ is the floor function and $\left\{ t\right\} 
 $ is the sawtooth function. Using the bounds $0\leq\left\{ t\right\} <1
 $ we get $$n\log\left(\log\left(n\right)\right)-\textrm{Li}\left(n\right)\leq\sum_{i=2}^{n}\log\left(\log\left(i\right)\right)\leq (n+1)\log\left(\log\left(n\right)\right)-\textrm{Li}\left(n\right)-\log\left(\log\left(2\right)\right)
 $$ where $\textrm{Li}\left(n\right)=\int_{2}^{n}\frac{1}{\log\left(t\right)}dt
 $ is the logarithmic integral. Finally $$\frac{\log^{n}\left(n\right)}{e^{\textrm{Li}\left(n\right)}}\leq\prod_{i=2}^{n}\log\left(i\right)\leq\frac{\log^{n+1}\left(n\right)}{e^{\textrm{Li}\left(n\right)}\log\left(2\right)}.
 $$
