> [!NOTE] Theorem > For all $n\in \mathbb{N},$ the $n$th [[Factorial|factorial]] can be approximated as follows $n!\sim \sqrt{ 2\pi n }\left( \frac{n}{e} \right)^{n}$where $e$ denotes [[Euler's Number|Euler's number]] and $\pi$ denotes [[Pi|pi]]. **Proof**: For all $n\in\mathbb{N},$ let $I_{n}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{n} x \, dx$By [[Integration by Parts]], $\begin{align} I_{n+2}&= [\cos^{n+1} x \sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} +(n+1) \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2}\cos^{n} x \, dx \\ &= (n+1) \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1-\sin^{2} \theta )\cos^{n} x \, dx \\ &= (n+1)(I_{n}-I_{n+2}) \end{align}$Thus $I_{n+2}=\frac{n+1}{n+2} I_{n}$Since $I_{0}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \, dx = \pi$ we have by induction that for all $n\geq 1$ $I_{2n}=\frac{\pi}{4^{n}} {2n \choose n}.$For all $k\geq 1,$ let $J_{k}=I_{k}I_{k-1}.$ Then for all $k \geq 2,$ $J_{k}=I_{k}I_{k-1}= \frac{k-1}{k} I_{k-1}I_{k-2} = \frac{k-1}{k} J_{k-1}.$Since $I_{1}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx=2$ we have $J_{1}=2\pi$ and hence by induction for all $k \geq 1,$ $J_{k}=\frac{2\pi}{k}.$Now $\sqrt{ \frac{n}{2} }I_{2n} =\sqrt{ \frac{n}{2} } \frac{\pi(2n)!}{4^{n} (n!)^{2}} = \frac{(2n)!}{e^{-2n}(2n)^{2n+1/2}}\left( \frac{e^{-n}n^{n+1/2}}{n!} \right)^{2}\pi\to \frac{\pi}{c} $as $n\to \infty.$ where $c=\lim_{ n \to \infty } \frac{n!}{e^{-n} n^{n+1/2}} $assuming the limit exists. For each $k\geq 1,$ we have $I_{k}\leq I_{k-1}$ and so for each $n\in \mathbb{N},$ $I_{2n}\leq \sqrt{ I_{2n}I_{2n-1} }=\sqrt{ J_{2n} }=\sqrt{ \frac{2\pi}{2n} }$and a similar argument gives $\sqrt{ \frac{2\pi}{2n+1}} \leq I_{2n}.$ From this it follows that, $\sqrt{ \frac{n}{2} }I_{n} \to \sqrt{ \frac{\pi}{2} }$as $n\to \infty.$ Hence we have $c=\sqrt{ 2\pi }$: that is, $\lim_{ n \to \infty } \frac{n!}{e^{-n} n^{n+1/2}}=2\pi$thus $n!\asymp \sqrt{2 \pi} e^{-n} n^{n+1 / 2}.$