Definite Integral from 0 to Half Pi of Even Power of Sine x

Theorem
Let $n \in \Z_{\ne 0}$ be a positive integer.

Then:
 * $\displaystyle \int_0^{\frac \pi 2} \sin^{2 n} x \ \mathrm d x = \dfrac {\left({2n}\right)!} {\left({2^n n!}\right)^2} \dfrac \pi 2$

Proof
Let $I_n = \displaystyle \int_0^{\frac \pi 2} \sin^n x \ \mathrm d x$.

Then: