A product formula for sines

Factorization. $X^n-1=\prod_{k=0}^{n-1}(X-\omega^k)$ with $\omega=e^{2i\pi/n}$.

Divide by $X-1$. $\frac{X^n-1}{X-1}=\prod_{k=1}^{n-1}(X-\omega^k)$.

Evaluate at $X=1$. $\lim_{X\to1}\frac{X^n-1}{X-1}=n$. And $\prod_{k=1}^{n-1}(1-\omega^k)=n$.

Modulus. $|1-\omega^k|^2=2-2\cos(2k\pi/n)=4\sin^2(k\pi/n)$, so $|1-\omega^k|=2\sin(k\pi/n)$ (positive for $k=1,\ldots,n-1$).

Conclusion. $\prod_{k=1}^{n-1}2\sin(k\pi/n)=n$, i.e. $\prod_{k=1}^{n-1}\sin(k\pi/n)=\frac{n}{2^{n-1}}$.