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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.

The Wallis integrals are $I_n={\int }_0^{\pi /2}{\sin }^{n}xdx$ for integers $n\ge 0$. Integration by parts gives $n{I}_n=(n-1)I_{n-2}$ for $n\ge 2$, with $I_0=\pi /2$, $I_1=1$. HenceBecause $0<\sin x<1$ in the integration interval, $0<I_n<I_{n-1}$ for $n\ge 1$. Combining this with the recurrence gives $2n/(2n+1)<I_{2n+1}/I_{2n}<1$, proving that the ratio tends to one and yielding the Wallis product.

The Wallis product isIts finite product is $W_n=2^{4n}(n!{)}^4/[(2n+1)((2n)!{)}^2]$. The Wallis integrals give $2W_n/\pi =I_{2n+1}/I_{2n}\to 1$, proving both convergence of the product and its value. The strict integral bounds also give $\pi n/(2n+1)<W_n<\pi /2$.