Stirling's Formula/Proof 2/Lemma 4

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

Then:
 * $\displaystyle \lim_{n \mathop \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$