# Finding the number of digits of a factorial

If the argument is not too big, this can be tackled with logarithms.

See that the number of digits of n! is ceil(log10(n!)) for n greater than 1.

By definition,

n! = n * (n - 1) * ... * 1


Therefore,

log(n!) = log(n) + log(n - 1) + ... + log(1)


Thus, a simple and fast solution in Python is:

def factorial_digit_count(n):
if n < 2:
return 1
return ceil(sum(log10(x) for x in range(1, n + 1)))


Using 64-bit IEEE binary floating point, this function returns correct results up to n = 3,121,515. This approximation evaluates to 18,916,606, rather than 18,916,607, which would be the correct answer according to Wolfram Alpha and my PARI program.