limits - Proof of $lim_{ntoinfty}frac{e^nn!}{n^nsqrt{n}}=sqrt{2pi}$

I'm looking for a proof of the following limit:

$$\lim_{n\to\infty}\frac{e^nn!}{n^n\sqrt{n}}=\sqrt{2\pi}$$
This follows from Stirling's Formula, but how can it be proven?

Answer

Existence of the Limit

The ratio of two consecutive terms is
$$\left.\frac{e^nn!}{n^{n+1/2}}\middle/\frac{e^{n-1}(n-1)!}{(n-1)^{n-1/2}}\right. =e\left(1-\frac1n\right)^{n-1/2}\tag{1}$$

Using the Taylor approximation
$$\log\left(1-\frac1n\right)=-\frac1n-\frac1{2n^2}-\frac1{3n^3}+O\left(\frac1{n^4}\right)\tag{2}$$
we can compute the logarithm of the right hand side of $(1)$:
$$1+\left(n-\frac12\right)\log\left(1-\frac1n\right) =-\frac1{12n^2}+O\left(\frac1{n^3}\right)\tag{3}$$
Therefore,

$$\begin{align} \lim_{n\to\infty}\frac{e^nn!}{n^{n+1/2}} &=e\prod_{n=2}^\infty\left(\frac{e^nn!}{n^{n+1/2}}\middle/\frac{e^{n-1}(n-1)!}{(n-1)^{n-1/2}}\right)\\ &=e\prod_{n=2}^\infty\left(e\left(1-\frac1n\right)^{n-1/2}\right)\\ &=\exp\left(1+\sum_{n=2}^\infty\left(1+\left(n-\frac12\right)\log\left(1-\frac1n\right)\right)\right)\\ &=\exp\left(1+\sum_{n=2}^\infty\left(-\frac1{12n^2}+O\left(\frac1{n^3}\right)\right)\right)\tag{4} \end{align}$$
Since the sum on the right hand side of $(4)$ converges, so does the limit on the left hand side.

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Log-Convexity of $\boldsymbol{\Gamma(x)}$

Since $\Gamma(x)$ is log-convex,
$$\begin{align} \sqrt{n}\,\Gamma(n+1/2) &\le\sqrt{n}\,\Gamma(n)^{1/2}\,\Gamma(n+1)^{1/2}\\ &=\Gamma(n+1)\tag{5} \end{align}$$
and
$$\begin{align} \Gamma(n+1) &\le\Gamma(n+1/2)^{1/2}\,\Gamma(n+3/2)^{1/2}\\ &=\sqrt{n+1/2}\,\Gamma(n+1/2)\tag{6} \end{align}$$
Therefore,
$$\sqrt{\frac{n}{n+1/2}}\le\sqrt{n}\frac{\Gamma(n+1/2)}{\Gamma(n+1)}\le1\tag{7}$$
Thus, by the Squeeze Theorem,
$$\lim_{n\to\infty}\sqrt{n}\frac{\Gamma(n+1/2)}{\Gamma(n+1)}=1\tag{8}$$

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Using the Recursion for $\boldsymbol{\Gamma(x)}$

Using the identity $\Gamma(x+1)=x\Gamma(x)$,
$$\begin{align} \sqrt{n}\color{#C00000}{\frac{(2n)!}{2^nn!}}\color{#00A000}{\frac1{2^nn!}} &=\sqrt{n}\frac{\color{#C00000}{1}}{\color{#00A000}{2}}\cdot\frac{\color{#C00000}{3}}{\color{#00A000}{4}}\cdot\frac{\color{#C00000}{5}}{\color{#00A000}{6}}\cdots\frac{\color{#C00000}{2n-1}}{\color{#00A000}{2n}}\\ &=\sqrt{n}\frac{\color{#C00000}{1/2}}{\color{#00A000}{1}}\cdot\frac{\color{#C00000}{3/2}}{\color{#00A000}{2}}\cdot\frac{\color{#C00000}{5/2}}{\color{#00A000}{3}}\cdots\frac{\color{#C00000}{n-1/2}}{\color{#00A000}{n}}\\ &=\sqrt{n}\color{#C00000}{\frac{\Gamma(n+1/2)}{\Gamma(1/2)}}\color{#00A000}{\frac{\Gamma(1)}{\Gamma(n+1)}}\tag{9} \end{align}$$

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Value of the Limit

Combining $(8)$ and $(9)$ yields
$$\begin{align} \lim_{n\to\infty}\left(\frac{e^nn!}{n^{n+1/2}}\right)^{-1} &=\lim_{n\to\infty}\frac{e^{2n}(2n)!}{(2n)^{2n+1/2}}\left(\frac{n^{n+1/2}}{e^nn!}\right)^2\\ &=\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt2\,4^n}\frac{(2n)!}{(n!)^2}\\ &=\lim_{n\to\infty}\frac1{\sqrt2}\frac{\Gamma(1)}{\Gamma(1/2)}\sqrt{n}\frac{\Gamma(n+1/2)}{\Gamma(n+1)}\\ &=\frac1{\sqrt{2\pi}}\tag{10} \end{align}$$
Therefore,
$$\lim_{n\to\infty}\frac{e^nn!}{n^{n+1/2}}=\sqrt{2\pi}\tag{11}$$