= Solution
Use <integration by parts> on $\sin^{n-1}x\sin x$, for $n\geq2$:
$$
\begin{aligned}
I_n
&=[-\sin^{n-1}x\cos x]_0^{\pi/2}
+(n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x\,dx\\
&=(n-1)(I_{n-2}-I_n).
\end{aligned}
$$
Both boundary terms are zero. Therefore the <Wallis integrals> satisfy
$$
\boxed{nI_n=(n-1)I_{n-2}.}
$$
Starting with $I_0=\pi/2$ and $I_1=1$, repeated application gives
$$
\begin{aligned}
I_{2n}&=\frac{1\cdot3\cdots(2n-1)}{2\cdot4\cdots2n}\frac\pi2
=\frac{(2n)!}{2^{2n}(n!)^2}\frac\pi2,\\
I_{2n+1}&=\frac{2\cdot4\cdots2n}{3\cdot5\cdots(2n+1)}
=\frac{2^{2n}(n!)^2}{(2n+1)!}.
\end{aligned}
$$
These formulas also hold at $n=0$ by empty products.
For $0<x<\pi/2$, $0<\sin x<1$, so $0<\sin^n x<\sin^{n-1}x$ when $n\geq1$. Integrating the strict inequalities proves $0<I_n<I_{n-1}$. In particular $I_{2n+1}/I_{2n}<1$. For $n\geq1$, the recurrence and $I_{2n-1}>I_{2n}$ give
$$
I_{2n+1}=\frac{2n}{2n+1}I_{2n-1}
>\frac{2n}{2n+1}I_{2n}.
$$
Thus
$$
\boxed{\frac{2n}{2n+1}<\frac{I_{2n+1}}{I_{2n}}<1.}
$$
At $n=0$ the same displayed bounds are $0<2/\pi<1$, also true directly from $I_1<I_0$.
The <squeeze theorem> now gives $I_{2n+1}/I_{2n}\to1$. Substituting the evaluated integrals shows
$$
\frac{I_{2n+1}}{I_{2n}}
=\frac1\pi\frac{2^{4n+1}(n!)^4}{(2n+1)((2n)!)^2},
$$
and hence
$$
\boxed{\lim_{n\to\infty}\frac{2^{4n+1}(n!)^4}{(2n+1)((2n)!)^2}=\pi.}
$$
This is twice the finite-product form of the <Wallis product>; the proof used only the integral recurrence and strict comparison of its integrands.
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