Wallis's Product/Original Proof

Proof
From the Reduction Formula for Integral of Power of Sine, we have:
 * $\ds (1): \quad \int \sin^n x \rd x = - \frac 1 n \sin^{n - 1} \cos x + \frac {n - 1} n \int \sin^{n - 2} x \rd x$

Let $I_n$ be defined as:
 * $\ds I_n = \int_0^{\pi / 2} \sin^n x \rd x$

As $\cos \dfrac \pi 2 = 0$ from Shape of Cosine Function, we have from $(1)$ that:
 * $(2): \quad I_n = \dfrac {n-1} n I_{n - 2}$

To start the ball rolling, we note that:
 * $\ds I_0 = \int_0^{\pi / 2} \rd x = \frac \pi 2$
 * $\ds I_1 = \int_0^{\pi / 2} \sin x \rd x = \bigintlimits {-\cos x} 0 {\pi / 2} = 1$

We need to separate the cases where the subscripts are even and odd:

By Shape of Sine Function, we have that on $0 \le x \le \dfrac \pi 2$:
 * $0 \le \sin x \le 1$

Therefore:
 * $0 \le \sin^{2 n + 2} x \le \sin^{2 n +1} x \le \sin^{2 n} x$

It follows from Relative Sizes of Definite Integrals that:
 * $\ds 0 < \int_0^{\pi / 2} \sin^{2 n + 2} x \rd x \le \int_0^{\pi / 2} \sin^{2 n + 1} x \rd x \le \int_0^{\pi / 2} \sin^{2 n} x \rd x$

That is:
 * $(3): \quad 0 < I_{2 n + 2} \le I_{2 n + 1} \le I_{2 n}$

By $(2)$ we have:
 * $\dfrac {I_{2 n + 2} } {I_{2 n} } = \dfrac {2 n + 1} {2 n + 2}$

Dividing $(3)$ through by $I_{2n}$ then, we have:
 * $\dfrac {2 n + 1} {2 n + 2} \le \dfrac {I_{2 n + 1}} {I_{2 n}} \le 1$

By Squeeze Theorem, it follows that:
 * $\dfrac {I_{2 n + 1} } {I_{2 n} } \to 1$ as $n \to \infty$

which is equivalent to:
 * $\dfrac {I_{2 n} } {I_{2 n + 1} } \to 1$ as $n \to \infty$

Now we take $(B)$ and divide it by $(A)$ to get:


 * $\dfrac {I_{2 n + 1} } {I_{2 n} } = \dfrac 2 1 \cdot \dfrac 2 3 \cdot \dfrac 4 3 \cdot \dfrac 4 5 \cdots \dfrac {2 n} {2 n - 1} \cdot \dfrac {2 n} {2 n + 1} \cdot \dfrac 2 \pi$

So:
 * $\dfrac \pi 2 = \dfrac 2 1 \cdot \dfrac 2 3 \cdot \dfrac 4 3 \cdot \dfrac 4 5 \cdots \dfrac {2 n} {2 n - 1} \cdot \dfrac {2 n} {2 n + 1} \cdot \paren {\dfrac {I_{2 n} } {I_{2 n + 1} } }$

Taking the limit as $n \to \infty$ gives the result.