Proving $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}=\sqrt{2 \pi }$

Elaborating on Daniel Schepler's comment: if $L = \lim_{n \to \infty} \frac{n!}{n^{n+1/2} e^{-n}}$ then $$\sqrt{\pi/2} = \lim_{n\to \infty} \frac{2^{2n} (n!)^2}{(2n)!\sqrt{2n+1}} = \lim_{n \to \infty} \frac{2^{2n}n^{2n+1} e^{-2n} L^2}{(2n)^{2n+1/2} e^{-2n} L \sqrt{2n+1}} = L\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{2}\sqrt{2n+1}} = \frac{L}{2}.$$

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For what it's worth, I tried to continue your approach by approximating $\log \left(\prod_{k=1}^n (1 - \frac{2k-1}{2n})\right)$ with the integral $n \int_0^{1-1/(2n)} \log(1-x) \, dx$ but it was terribly messy and didn't seem tight enough to get the exact equivalence with $\sqrt{2} e^{-n}$. But I think the above approach is the intended approach.