$\prod _{k=2}^{n} {\log k}$ is big-$O$ of what?

Question

$$\prod _{k=2}^{n} {\log k}$$

is a big-$O$ of what?

I can see it $O(n!)$ but is there a tighter solution?

Answer 1

If your product is $P(n)$, then $\log P(n) = \sum_{k=2}^n \log \log k \le n \log \log n$, so $P(n) \le \exp((n-1) \log \log n) = (\log n)^{n-1}$.

Answer 2

Since $\log n \leq n^\alpha$, for $\alpha>0$ and $n>N(\alpha)$, you can say

$$ P(n)\leq \prod_{k=2}^n k^\alpha = (n!)^\alpha, \mbox{ for } n>N(\alpha) $$

This holds for every positive alpha, although the index $N$ where this bound starts to be true is pushed forward as $\alpha\to 0$.