The product of sin(kπ/n)

Factorisation. The $n$-th roots of unity other than $1$ are the roots of $x^{n-1}+\dots+x+1=\prod_{k=1}^{n-1}(x-\omega^k)$. At $x=1$: $n=\prod_{k=1}^{n-1}(1-\omega^k)$.

Moduli. For $\theta\in(0,2\pi)$,
$$|1-e^{i\theta}|=2\sin\frac\theta2.$$
With $\theta=\frac{2\pi k}n$ this is $2\sin\frac{k\pi}n>0$. Taking moduli in the product,
$$n=\prod_{k=1}^{n-1}2\sin\frac{k\pi}n=2^{n-1}\prod_{k=1}^{n-1}\sin\frac{k\pi}n,$$
which is the claim.

Check. $n=3$: $\sin\frac\pi3\sin\frac{2\pi}3=\frac34=\frac3{2^2}$ ✓. $n=4$: $\frac{\sqrt2}2\cdot1\cdot\frac{\sqrt2}2=\frac12=\frac4{2^3}$ ✓.

Lesson. A trigonometric product over equally spaced angles often comes from evaluating $\prod(x-\omega^k)$ at a well-chosen point.