What is the Big-$O$ of $\prod_{i=2}^n log{(i)}$ As I can see 
$$\prod_{i=2}^n log{(i)} = log(2) \cdot log(3) \cdot log(4)...log(p) \cdot log(p+1) ... log(n)= O(k^n)$$
when
$$ log(p) = k$$
Am I right? 
Is there any better Big-$O$ for it?
 A: \begin{align*}\log\left(\prod_{i=2}^{n}\log(i)\right)&=\sum_{i=2}^{n}\log\log(i)\\\text{(Jensen's Inequality)}&\leq (n-1)\log\left(\frac{1}{n-1}\sum_{k=2}^{n}\log(i)\right)\\\text{(again)}&\leq(n-1)\log\log\left(\frac{1}{n-1}\sum_{k=2}^{n}i\right)\\&=(n-1)\log\log\left(\frac{n+2}{2}\right)
\end{align*}
This gives $\prod_{i=2}^{n}\log(i)\leq \left(\log\left(\frac{n+2}{2}\right)\right)^{n-1}.$
A: Here's my estimation for the product in the $n\rightarrow\infty$ limit. Take the log is the first step. Then
$$\log\left(\prod_{i=2}^n\log i\right)=\sum_{i=2}^n\log\log i.$$
Then using the concavity of $\,\log\log x$, we know (assuming natural log)
$$\sum_{i=2}^n\log\log i\geq\left.\int_{\frac{3}{2}}^{n+\frac{1}{2}}\log\log x\,dx=x\log\log x\,\right|_{\frac{3}{2}}^{n+\frac{1}{2}}-\int_{\frac{3}{2}}^{n+\frac{1}{2}}\frac{dx}{\log x}.$$
The second part gives the logarithmic integral, whose full asymptotic expansion can be found on wikipedia. The error can be estimated to be a constant, because the leading-order error is given by the difference of areas of two "triangles" (approximately) in $[i-\frac{1}{2},i+\frac{1}{2}]$ summed over $i$, i.e.,
$$\sum_{i=2}^\infty\frac{1}{2}|(\log\log x)''|_{x=i}\left(\frac{1}{2}\right)^2\times 2=\frac{1}{4}\sum_{i=2}^\infty\frac{\log i+1}{i^2\log^2i}<\infty$$
is convergent. This means the integral estimator is off by a constant at most. But the logarithmic integral is not an elementary function.
