TITLE: Stirling approximation note
QUESTION [8 upvotes]: During my study to Stirling approximation I find this formula  $n! \approx \sqrt{2\pi  n}  n^{n}e^{-n} $ but we know that  $ 0!  =1 $ And in this formula if we replace every  $ n $ with  $ 0$ we will have $ 0^0$ which is undefined.  My question is how we get  $0! $ by using this formula? ..  can anyone explain that for me.  I'll appreciate that..  thanks

REPLY [2 votes]: The relation $n! \approx \sqrt{2\pi  n}\  n^n e^{-n}$ means $\lim\limits_{n\to\infty} \dfrac{\sqrt{2\pi n}\  n^n e^{-n}}{n!} = 1\vphantom{\frac{\int}{\displaystyle\sum}}$.  That means the ratio can be made as close to $1$ as desired by making $n$ big enough.  So nothing depends on what happens with small values of $n$.