Number of Digits in Factorial

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

The number of digits in $n!$ is approximately:
 * $1 + \floor {\dfrac 1 2 \paren {\log_{10} 2 + \log_{10} \pi} + \dfrac 1 2 \log_{10} n + n \paren {\log_{10} n - \log_{10} e} }$

when $n!$ is shown in decimal notation.

This evaluates to:
 * $1 + \floor {\paren {n + \dfrac 1 2} \log_{10} n - 0.43429 \ 4481 \, n + 0.39908 \ 9934}$

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

from which the result can be calculated.

To count the number of digits:

We have:
 * Common Logarithm of $2$: $\log_{10} 2 \approx 0.30102 \ 9996$
 * Common Logarithm of $\pi$: $\log_{10} \pi \approx 0.49714 \ 9873$
 * Common Logarithm of $e$: $\log_{10} e \approx 0.43429 \ 4481$

Hence:

Hence the result from Number of Digits in Number.