Number of Digits in Factorial

Theorem
Let $n!$ denote the factorial of $n$.

The number of digits in $n!$ is approximately:
 * $1 + \left\lfloor{\dfrac 1 2 \left({\log_{10} 2 + \log_{10} \pi}\right) + \dfrac 1 2 \log_{10} n + n \left({\log_{10} n - \log_{10} e}\right)}\right\rfloor$

when $n!$ is shown in decimal notation.

This evaluates to:
 * $1 + \left\lfloor{\left({n + \dfrac 1 2}\right) \log_{10} n - 0.43429 \ 4481 \, n + 0.39908 \ 9934}\right\rfloor$

Proof
From Stirling's Formula:
 * $n! \sim \sqrt {2 \pi n} \left({\dfrac n e}\right)^n$

from which the result can be calculated.

To count the number of digits:

We have:
 * Common Logarithm of $2$: $\ln 2 \approx 0.30102 \ 9996$
 * Common Logarithm of $\pi$: $\ln \pi \approx 0.49714 \ 9873$
 * Common Logarithm of $e$: $\ln e \approx 0.43429 \ 4481$

Hence:

Hence the result from Number of Digits in Number.