Factorial number of digits Is there any neat way to solve how many digits the number $20!$ have? I'm looking a solution which does not use computers, calculators nor log tables, just pen and paper.

I will use log as the base b logarithm and ln as the natural log. 
Then number of digits of x in base b is given by one more than the floor of the log(x). 
log(n!)=sum(log(k)) for k=1,2,...,n
We can interpret this as the Riemann sum for the integral from 1 to n of log(x) dx. This integral is actually a lower bound. The upper bound is the same integral from 2 to n+1 rather than 1 to n. 
The lower bound integral is given by n log(n)-(n-1)/ln(b).  The upper bound gives (n+1) log(n+1)-n/ln(b) -2ln(2)+1/ln(b). 
For n=20, working in base 10, we get about 17.8 as the lower bound and 18.9 as the upper bound. One more than the floor gives 18 or 19 digits. Not surprisingly, the answer is 19 as the lower bound is nearly 19 digits and the upper is nearly 20. 
The Riemann sum approximations will get better as n increases, but the answer is already quite good by n=20. 
