 Solution

Use integration by parts on ${\sin }^{n-1}x\sin x$, for $n\ge 2$:Both boundary terms are zero. Therefore the Wallis integrals satisfyStarting with $I_0=\pi /2$ and $I_1=1$, repeated application givesThese 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 $n\ge 1$. Integrating the strict inequalities proves $0<I_n<I_{n-1}$. In particular $I_{2n+1}/I_{2n}<1$. For $n\ge 1$, the recurrence and $I_{2n-1}>I_{2n}$ giveThusAt $n=0$ the same displayed bounds are $0<2/\pi <1$, also true directly from $I_1<I_0$.

The squeeze theorem now gives $I_{2n+1}/I_{2n}\to 1$. Substituting the evaluated integrals showsand henceThis is twice the finite-product form of the Wallis product; the proof used only the integral recurrence and strict comparison of its integrands.