TITLE: Estimate number of digits of 333!
QUESTION [5 upvotes]: How do I estimate the number of digits of 333! in a rigorous way? 
What I did so far is just counting in the following manner: 
1 digit for the first 99 numbers and then 2 digits for the remaining 234 numbers of 333!. This sums up to approx. 567 digits. Which is very rough compared to 698 digits which is the real solution.
Any other suggestions?

REPLY [2 votes]: For any (large) $N$ we have
$$\log(N!)=\sum_{k=1}^N\log k={1\over\ln10}\sum_{k=1}^N\ln k={1\over\ln10}\sum_{k=1}^N(\ln(k/N)+\ln N)={N\over\ln 10}\left(\ln N+{1\over N}\sum_{k=1}^N\ln(k/N)\right)\approx{N\over\ln10}\left(\ln N+\int_0^1\ln x\,dx\right)={N\over\ln10}(\ln N-1)$$
For $N=333$, this gives
$$\log(333!)\approx695.35$$
which suggests there are approximately $\lceil695.35\rceil=696$ digits in $333!$.
To be more rigorous about the integral approximation, we have
$$-1=\int_0^1\ln x\,dx\lt{1\over N}\sum_{k=1}^N\ln(k/N)\lt\int_{1/N}^1\ln x\,dx={\ln N+1\over N}-1$$
so
$${N\over\ln10}(\ln N-1)\lt\log(N!)\lt{N\over\ln10}(\ln N-1)+{\ln N+1\over\ln10}$$
which gives the following range for the number of digits:
$$696\le\lceil\log(333!)\rceil\le699$$