From $\displaystyle \sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ follows that $\displaystyle 2i\sin(\theta)e^{i\theta}=e^{i2\theta}-1$.Show that $\displaystyle \displaystyle{ \int_0^\pi \sin^{2n}(\theta)d\theta=\pi\frac{(2n)!}{2^{2n}(n! )^2} }$.

Exponentiating by $\displaystyle 2n$ gets, using $\displaystyle \theta$ and $\displaystyle -\theta$, respectively:

1.$\displaystyle \displaystyle{ 2^{2n}i^{2n}\sin^{2n}(\theta)e^{i2n\theta}=(e^{i2\ theta}-1)^{2n} }$

2.$\displaystyle \displaystyle{ 2^{2n}i^{2n}\sin^{2n}(\theta)e^{-i2n\theta}=(e^{-i2\theta}-1)^{2n} }$.

Subtracting2.from1.:

$\displaystyle \displaystyle{ 2^{2n}i^{2n}\sin^{2n}(\theta)\left[\underbrace{e^{in\theta}-e^{-in\theta}}_{2i\sin(2n\theta)}\right]=(e^{i2\theta}-1)^{2n}-(e^{-i2\theta}-1)^{2n} }$.

Integrating both sides from $\displaystyle 0$ to $\displaystyle \pi$:$\displaystyle

2^{2n+1}i^{2n+1}\int_0^\pi \sin^{2n}(\theta)\sin(2n\theta)d\theta=\int_0^\pi\ left\{ (e^{i2\theta}-1)^{2n}-(e^{-i2\theta}-1)^{2n} \right\} \stackrel{?}{=}0$

What to do with $\displaystyle \sin(2n\theta)$ inside integral, and, is RHS really zero? Assuming everything is OK before that.

$\displaystyle ?$: Two leading $\displaystyle 1$'s cancel and all other terms in expansion would give $\displaystyle \displaystyle{ \int_0^\pi e^{i2k\theta}d\theta=0 \text{ ; } k=\pm 1, \pm 2, \hdots, \pm 2n }$