<s>[INST] Prove Stirling's formula given that for $I_n = \int_0^{\pi/2} \sin^n\theta \, d\theta$ we have $I_{2n+1}/I_{2n} \rightarrow 1$ For $I_n = \int_0^{\pi/2} \sin^n\theta \, d\theta$ it is possible to show (using integration by parts and $\sin^2\theta = 1-\cos^2\theta$) that: $nI_n = (n-1)I_{n-2}$
We can then repeatedly apply this relation to $\frac{I_{2n+1}}{I_{2n}}$ to show:
$$\frac{I_{2n+1}}{I_{2n}} = \frac{2^{4n+1}(n!)^4}{\pi (2n)!(2n+1)!}$$
This ratio converges to $1$ since $I_{2n+2}/I_{2n} \rightarrow 1$ (easy to show with the above recursive relation)  which is always larger than $I_{2n+1}/I_{2n}$. Since $I_{2n}/I_{2n} = 1$, we can sandwich the desired ratio.
I also know that (from Show that $n!e^n/n^{n+1/2} \leq e^{1/(4n)}C$) that: $r_n = n!e^n/n^{n+1/2} \leq e^{1/(4n)}C$ for all $n$ and for $C = \lim_{n\rightarrow \infty} n!e^n/n^{n+1/2}$. Now I want to show Stirling's formula:
$$n! \sim \sqrt{2\pi}n^{n+1/2}/e^n$$
What I've tried
Using the upper bound on $r_n$ it is clear that all that remains is to show that $C = \sqrt{2\pi}$. I have seen trick on related problems involving expressing the integral $I_n$ in terms of the beta function, but I think another trick is needed as well (perhaps one involving introducing a term $e^{i\theta}$ but I am struggling to get anything more specific than that.
 [/INST] Note that you know a little bit more from the linked question - that $n!/(n^{n+1/2} e^{-n})$ decreases to $C$.
Using this and the upper bound you have, note that $$ \frac{I_{2n+1}}{I_{2n}} \le \frac{2^{4n + 1} C^4 n^{4n + 2} e^{-4n} e^{1/n}}{\pi \cdot C(2n)^{2n+1/2}e^{-2n} \cdot C(2n+1)^{2n + 3/2} e^{-2n - 1}} \\ = \frac{C^2}{2\pi} \frac{e^{1+1/n}}{(1 + 1/2n)^{2n + 3/2}} =: a_n$$
Since $I_{2n+1}/I_{2n} \to 1,$ we can conclude that $\liminf a_n \ge 1$. But $\liminf a_n = \frac{C^2}{2\pi}$, thus telling us that $C \ge \sqrt{2\pi}$.
Similarly, we can develop the lower bound $$ \frac{I_{2n+1}}{I_{2n}} \ge \frac{C^2}{2\pi} \frac{e}{e^{1/8n + 1/8n + 4} (1+1/2n)^{2n + 3/2}} =: b_n, $$
and argue that $\frac{C^2}{2\pi} =  \limsup b_n \le 1,$ ergo $C \le \sqrt{2\pi}$.
Thus we have shown that $\sqrt{2\pi} \le C \le \sqrt{2\pi},$ which of course implies that $C = \sqrt{2\pi}$.
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