Stirling's formula gives an approximation for $n!$ , the factorial function. It is

$$ n! \approx \sqrt{2n\pi} n^n e^{-n} $$

We can derive this from the gamma function. Note that for large $x$ ,

\begin{equation} \label{eq:glx} \Gamma(x) = \sqrt{2\pi}x^{x-\frac{1}{2}} e^{-x + \mu(x)} \end{equation} where

$$ \mu(x) = \sum_{n=0}^\infty \left(x+ n +\frac{1}{2}\right) \ln \left(1+ \frac{1}{x+n} \right) -1 = \frac{\theta}{12x} $$

with $0 < \theta < 1$ . Taking $x=n$ and multiplying by $n$ , we have

\begin{equation} \label{eq:penult} n! = \sqrt{2\pi} n^{n+\frac{1}{2}} e^{-n + \frac{\theta}{12n}} \end{equation} Taking the approximation for large $n$ gives us Stirling's formula.

There is also a big-O notation version of Stirling's approximation:

\begin{equation} \label{eq:bigover} n! = \left(\sqrt{2\pi n}\right)\left(\frac{n}{e}\right)^n \left(1+\mathcal{O}\left(\frac{1}{n}\right)\right) \end{equation} We can prove this equality starting from (). It is clear that the big-O portion of () must come from $e^{\frac{\theta}{12n}}$ , so we must consider the asymptotic behavior of $e$ .

First we observe that the Taylor series for $e^x$ is

$$ e^x = 1 + \frac{x}{1} + \frac{x^2}{2!} + \frac {x^3}{3!} + \cdots $$

But in our case we have $e$ to a vanishing exponent. Note that if we vary $x$ as $\frac{1}{n}$ , we have as $ n \longrightarrow \infty $

$$ e^x = 1 + \mathcal{O}\left(\frac{1}{n}\right) $$

We can then (almost) directly plug this in to () to get () (note that the factor of 12 gets absorbed by the big-O notation.)