Approach me 3

The problem can be solved by hand also.

Number of digits in a number $n$ is equal to $\lfloor log_{10} n \rfloor +1$ .

Here we need $\lfloor log_{10}(1000!)\rfloor +1 = 1+ \lfloor \sum_{i=1}^{1000}log_{10} i \rfloor$

This gives the answer as $1+ \lfloor 2567.6046 \rfloor = \boxed{2568}$

The number of power to 10 is: $log_10 (1000! = M * log(1000)$ with $M = \frac{1}{log 10}$

Applying Stirling's formular one gets: $log_{10}(1000!) = M *(0.5 log 2*\pi +0.5 * 1000*log 1000- 1000 + R_n$ that equals roughly $2567.6046 + M * R_n$

with $R_n \le 12000^{-1}$ , therefore 1000! actually has 2568 digits