Proving an inequality involving reduction sine I'm trying to prove that $I_{2n + 2} \leqslant I_{2n + 1} \leqslant I_{2n}$ where
$$
I_k = \int_0^{\pi/2} \sin^k x\, dx = \frac {k - 1} {k}\int_0^{\pi/2} \sin^{k - 2} x \, dx 
$$
From this I have worked out that
$$
I_{2n + 2} = \frac{2n + 1} {2n + 2} \bullet  \frac{2n - 1} {2n} \bullet  \frac{2n - 3} {2n - 2} \bullet \dotsi \bullet \frac{5} {6} \bullet   \frac{3} {4} \bullet   \frac{1} {2} \bullet   \frac{\pi} {2}
$$
$$
I_{2n + 1} = \frac{2n} {2n + 1} \bullet  \frac{2n - 2} {2n - 1} \bullet  \frac{2n - 4} {2n - 3} \bullet \dotsi \bullet \frac{6} {7} \bullet \frac{4} {5} \bullet \frac{2} {3}
$$
$$
I_{2n} = \frac{2n - 1} {2n} \bullet  \frac{2n - 3} {2n - 2} \bullet  \frac{2n - 5} {2n - 4} \bullet \dotsi \bullet \frac{5} {6} \bullet \frac{3} {4} \bullet \frac{1} {2} \bullet \frac{\pi} {2}
$$
Then I found that
$$
\frac{2n + 1} {2n + 2} I_{2n} \leqslant \frac{2n} {2n + 1} \bullet  \frac{2n - 2} {2n - 1} \bullet  \dotsi \bullet \frac{6} {7} \bullet \frac{4} {5} \bullet \frac{2} {3} \leqslant I_{2n}
$$
And from there I haven't worked out anything conclusive besides dividing everything by $I_{2n}$ which doesn't help me that much. So could anyone tell me what to do next or maybe point out a mistake?

The reduction formula is not needed. On our interval $0\le \sin^{k+1} x\le \sin^k x$. The result follows.
