Stirling's Formula/Refinement

Theorem
A refinement of Stirling's Formula is:
 * $n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + \dfrac 1 {12 n} }$

where $\sim$ denotes asymptotically equal.

Proof
From Limit of Error in Stirling's Formula:


 * $e^{1 / \paren {12 n + 1} } \le \dfrac {n!} {\sqrt {2 \pi n} n^n e^{-n} } \le e^{1 / 12 n}$