Stirling’s approximation
\[\begin{eqnarray} \ln n! & = & \sum_{i=1}^{n} \ln i \nonumber \\ & \le & \int_{1}^{n} \ln x \ dx \nonumber \\ & = & n \ln n - (n - 1) \nonumber \end{eqnarray} .\]Theorem
\[\begin{equation} n! = \sqrt{2 \pi n} \left( \frac{n}{e} \right)^{n} \left( 1 + O \left( \frac{1}{n} \right) \right) \end{equation}\]proof
\[\begin{eqnarray} \ln n! & = & \sum_{i=1}^{n} \ln i \nonumber \\ \int_{1}^{n} \ln x \ dx & = & n \ln n - (n - 1) \nonumber \end{eqnarray} .\] \[\begin{eqnarray} f(i) & := & \ln i \nonumber \\ k > 0, \ f^{(k)}(i) & = & (-1)^{k + 1} \frac{(k - 1)!}{i^{k}} \nonumber \\ \ln n! & = & \sum_{i=1}^{n} f(i) \nonumber \end{eqnarray} .\] \[\begin{eqnarray} \sum_{i=1}^{n} \ln i - (n \ln n - (n -1)) & = & \sum_{i=1}^{n} \ln i - n \ln n + n - 1 \nonumber \\ & = & \sum_{i=1}^{n} f(i) - \int_{1}^{n} f(x) \ dx \nonumber \\ & = & \sum_{k=1}^{p} \frac{B_{k}}{k!} \left( f^{(k-1)}(n) - f^{(k-1)}(1) \right) + R_{n, p} \quad (\because \text{Euler-Maclaurin formula}) \nonumber \\ & = & B_{1} ( \ln n - \ln 1 ) + \sum_{k=2}^{p} \frac{B_{k}}{k!} \left( (-1)^{k} \frac{(k - 2)!}{n^{k-1}} - (-1)^{k} (k - 2)! 1 \right) + R_{n, p} \nonumber \\ & = & \frac{1}{2} \ln n + \sum_{k=2}^{p} (-1)^{-1} \frac{B_{k}}{(k - 1)k} \left( \frac{1}{n^{k-1}} - 1 \right) + R_{n, p} \label{equation_euler_maclaurin_formula} \end{eqnarray}\]where $B_{k}$ is Bernoulli number (especially $B_{1}=1/2$) and $R_{n, p}$ is $p$-th remainder term in Euler-Maclaurin formula.
Arranging the above equation and taking the limit as $n \rightarrow \infty$,
\[y := \lim_{n \rightarrow \infty} \left( \sum_{i=1}^{n} \ln i - n \ln n + n - \frac{1}{2} \ln n \right) = 1 - \sum_{k=2}^{p} (-1)^{k} \frac{B_{k}}{(k - 1)k} + \lim_{n \rightarrow \infty} R_{n, p} .\]By Euler-Maclaurin formula,
\[\begin{eqnarray} R_{n, p} & = & \lim_{n^{\prime} \rightarrow \infty} R_{n^{\prime}, p} + O \left( \frac{1}{n^{p}} \right) \quad (\because \text{Euler-Maclaurin formula}) \nonumber \\ & = & y - 1 + \sum_{k=2}^{p} (-1)^{k} \frac{B_{k}}{(k - 1)k} + O \left( \frac{1}{n^{p}} \right) \nonumber \end{eqnarray}\]Hence \(\eqref{equation_euler_maclaurin_formula}\)
\[\begin{eqnarray} & & \sum_{i=1}^{n} \ln i - n \ln n + n - 1 = \frac{1}{2} \ln n + \sum_{k=2}^{p} (-1)^{-1} \frac{B_{k}}{(k - 1)k} \left( \frac{1}{n^{k-1}} - 1 \right) + R_{n, p} \nonumber \\ & \Leftrightarrow & \sum_{i=1}^{n} \ln i - n \ln n + n \ln e = \frac{1}{2} \ln n + \sum_{k=2}^{p} (-1)^{-1} \frac{B_{k}}{(k - 1)k} \left( \frac{1}{n^{k-1}} - 1 \right) + y + \sum_{k=2}^{p} (-1)^{k} \frac{B_{k}}{(k - 1)k} + O \left( \frac{1}{n^{p}} \right) \nonumber \\ & \Leftrightarrow & \sum_{i=1}^{n} \ln i + n \ln \left( \frac{e}{n} \right) = \frac{1}{2} \ln n + \sum_{k=2}^{p} (-1)^{-1} \frac{B_{k}}{(k - 1)k} \frac{1}{n^{k-1}} + y + O \left( \frac{1}{n^{p}} \right) \nonumber \\ & \Leftrightarrow & \sum_{i=1}^{n} \ln i = n \ln \left( \frac{n}{e} \right) + \frac{1}{2} \ln n + \sum_{k=2}^{p} (-1)^{-1} \frac{B_{k}}{(k - 1)k} \frac{1}{n^{k-1}} + y + O \left( \frac{1}{n^{p}} \right) \end{eqnarray}\]Taking the exponential of both sides,
\[\begin{eqnarray} & & n! = \left( \frac{n}{e} \right)^{n} n^{1/2} \exp \left( \sum_{k=2}^{p} (-1)^{-1} \frac{B_{k}}{(k - 1)k} \frac{1}{n^{k-1}} \right) e^{y} \exp \left( O \left( \frac{1}{n^{p}} \right) \right) \nonumber \\ & \Leftrightarrow & n! = \left( \frac{n}{e} \right)^{n} n^{1/2} \exp \left( \sum_{k=2}^{p} (-1)^{-1} \frac{B_{k}}{(k - 1)k} \frac{1}{n^{k-1}} \right) e^{y} \exp \left( O \left( \frac{1}{n^{p}} \right) \right) \end{eqnarray}\]Let $p=1$. By Wallis’ product, $e^{y} = \sqrt{2\pi}$.
\[\begin{eqnarray} n! & = & \left( \frac{n}{e} \right)^{n} n^{1/2} \sqrt{2\pi} \exp \left( O \left( \frac{1}{n} \right) \right) \nonumber \\ & = & \left( \frac{n}{e} \right)^{n} \sqrt{2\pi n} \exp \left( O \left( \frac{1}{n} \right) \right) \end{eqnarray}\]TODO
$\Box$