Stirling's Formula normalization An obvious way to get estimates on $n!$ is to compare $\sum\log k$ to $\int\log t$. If one could get Stirling's formula this way that  would strike me as the "right" proof, because it would be clear why it works.
This morning I came much closer to this than I have in the past; in fact fairly straightforward comparisons of sums to integrals show that  $$n!\sim c\sqrt n\left(\frac ne\right)^n.$$
Question: I wonder if there's some cheap trick to show that if $n!\sim c\sqrt n(n/e)^n$ then $c=\sqrt{2\pi}$.
 A: You could calculate the normalization for the Gaussian approximation to the binomial distribution.
A: A Cheap Method
As a function of $k$, $\frac{n^k}{k!}$ reaches its maximum when $k=n-1$ and $k=n$. Furthermore,
$$
\begin{align}
\log\left(\frac{n^k}{k!}\frac{n!}{n^n}\right)
&\stackrel{k\lt n}{=}\log\left[\left(1-\frac0n\right)\left(1-\frac1n\right)\cdots\left(1-\frac{n-k-1}n\right)\right]\\
&\stackrel{k\gt n}{=}\log\left[\frac1{1+\frac1n}\frac1{1+\frac2n}\cdots\frac1{1+\frac{k-n}n}\right]
\end{align}
$$
Both of these give a second order difference asymptotic to $-\frac1n$.
Therefore, applying Laplace's Method, we get
$$
\begin{align}
e^n\frac{n!}{n^n}
&=\sum_{k=0}^\infty\frac{n^k}{k!}\frac{n!}{n^n}\\
&\sim\int_{-\infty}^\infty e^{-\frac{x^2}{2n}}\,\mathrm{d}x\\[9pt]
&=\sqrt{2\pi n}
\end{align}
$$

A More Detailed Method
For $k\lt n$,
$$
\begin{align}
&\log\left(\frac{n^k}{k!}\right)-\log\left(\frac{n^n}{n!}\right)\\
&=\log\left(\frac{n-0}{n}\right)+\log\left(\frac{n-1}{n}\right)+\cdots+\log\left(\frac{n-(n-k-1)}{n}\right)\\
&=\sum_{j=0}^{n-k-1}\log\left(1-\frac jn\right)\tag{1a}
\end{align}
$$
and, bounding a Riemann Sum by integrals,
$$
\int_0^{\frac{n-k}n}\log(1-t)\,\mathrm{d}t
\le\sum_{j=0}^{n-k-1}\log\left(1-\frac jn\right)\frac1n
\le\int_0^{\frac{n-k-1}n}\log(1-t)\,\mathrm{d}t\tag{1b}
$$
For $x\in[0,1]$, $-\frac12x^2-\frac12x^3\le\int_0^x\log(1-t)\,\mathrm{d}t\le-\frac12x^2$. Thus,
$$
-\frac12\left(\frac{n-k}{\sqrt{n}}\right)^2-\frac1{2\sqrt{n}}\left(\frac{n-k}{\sqrt{n}}\right)^3
\le\sum_{j=0}^{n-k-1}\log\left(1-\frac jn\right)
\le-\frac12\left(\frac{n-k-1}{\sqrt{n}}\right)^2\tag{1c}
$$
Exponentiating and dividing by $\sqrt{n}$ gives
$$
\exp\left(\scriptsize{-\frac12\left(\frac{n-k}{\sqrt{n}}\right)^2-\frac1{2\sqrt{n}}\left(\frac{n-k}{\sqrt{n}}\right)^3}\right)\frac1{\sqrt{n}}
\le\frac{n^k}{k!}\frac{n!}{n^n}\frac1{\sqrt{n}}
\le\exp\left(\scriptsize{-\frac12\left(\frac{n-k-1}{\sqrt{n}}\right)^2}\right)\frac1{\sqrt{n}}\tag{1d}
$$
Summing over $k\in[0,n-1]$ and applying the Squeeze Theorem as $n\to\infty$ yields
$$
\lim_{n\to\infty}\frac{n!}{n^n}\frac1{\sqrt{n}}\sum_{k=0}^{n-1}\frac{n^k}{k!}
=\int_0^\infty e^{-t^2/2}\,\mathrm{d}t\tag{1e}
$$

For $k\ge n$,
$$
\begin{align}
&\log\left(\frac{n^k}{k!}\right)-\log\left(\frac{n^n}{n!}\right)\\
&=\log\left(\frac{n}{n+1}\right)+\log\left(\frac{n}{n+2}\right)+\cdots+\log\left(\frac{n}{n+(k-n)}\right)\\
&=-\sum_{j=1}^{k-n}\log\left(1+\frac jn\right)\tag{2a}
\end{align}
$$
and, bounding a Riemann Sum by integrals,
$$
-\int_0^{\frac{k-n+1}n}\log(1+t)\,\mathrm{d}t
\le-\sum_{j=1}^{k-n}\log\left(1+\frac jn\right)\frac1n
\le-\int_0^{\frac{k-n}n}\log(1+t)\,\mathrm{d}t\tag{2b}
$$
For $x\in[0,1]$, $-\frac12x^2\le-\int_0^x\log(1+t)\,\mathrm{d}t\le-\frac12x^2+\frac16x^3$. Thus,
$$
-\frac12\left(\frac{k-n+1}{\sqrt{n}}\right)^2
\le-\sum_{j=1}^{k-n}\log\left(1+\frac jn\right)
\le-\frac12\left(\frac{k-n}{\sqrt{n}}\right)^2+\frac1{6\sqrt{n}}\left(\frac{k-n}{\sqrt{n}}\right)^3\tag{2c}
$$
Exponentiating and dividing by $\sqrt{n}$ yields
$$
\exp\left(\scriptsize{-\frac12\left(\frac{k-n+1}{\sqrt{n}}\right)^2}\right)\frac1{\sqrt{n}}
\le\frac{n^k}{k!}\frac{n!}{n^n}\frac1{\sqrt{n}}
\le\exp\left(\scriptsize{-\frac12\left(\frac{k-n}{\sqrt{n}}\right)^2+\frac1{6\sqrt{n}}\left(\frac{k-n}{\sqrt{n}}\right)^3}\right)\frac1{\sqrt{n}}\tag{2d}
$$
Summing over $k\in[n,2n]$ and applying the Squeeze Theorem as $n\to\infty$ yields
$$
\lim_{n\to\infty}\frac{n!}{n^n}\frac1{\sqrt{n}}\sum_{k=n}^{2n}\frac{n^k}{k!}
=\int_0^\infty e^{-t^2/2}\,\mathrm{d}t\tag{2e}
$$

For $t\ge1$, $\log\left(\frac1{1+t}\right)\le-\log(2)$. Using this and $(2b)$, we get, for $k\ge2n$,
$$
\begin{align}
-\sum_{j=1}^{k-n}\log\left(1+\frac jn\right)
&\le-\sum_{j=1}^n\log\left(1+\frac jn\right)
-\sum_{j=n+1}^{k-n}\log(2)\\
&\le(1-2\log(2))n-\log(2)(k-2n)\tag{3a}
\end{align}
$$
Thus, for $k\ge2n$,
$$
\frac{n^k}{k!}\le\frac{n^n}{n!}\left(\frac e4\right)^n2^{-(k-2n)}\tag{3b}
$$
Therefore,
$$
\lim_{n\to\infty}\frac{n!}{n^n}\frac1{\sqrt{n}}\sum_{k\gt2n}\frac{n^k}{k!}
\le\lim_{n\to\infty}\frac1{\sqrt{n}}\left(\frac e4\right)^n
=0\tag{3c}
$$

Combining $\text{(1e)}$, $\text{(2e)}$, and $\text{(3c)}$, we get
$$
\begin{align}
\lim_{n\to\infty}\frac{n!}{n^n}\frac{e^n}{\sqrt{n}}
&=\lim_{n\to\infty}\frac{n!}{n^n}\frac1{\sqrt{n}}\sum_{k=0}^\infty\frac{n^k}{k!}\\
&=2\int_0^\infty e^{-x^2/2}\,\mathrm{d}x\\[9pt]
&=\sqrt{2\pi}\tag4
\end{align}
$$
