Let $Z=e^{i\frac{2\pi}{n}}$. Then $Z$, $Z^2$, $\dots$, $Z^{n-1}$ are $(n-1)$ complex zeros of $z^n-1=0$, or the zeros of $z^{n-1} + z^{n-2}+ \cdots + z + 1 = 0$. This implies: $$z^{n-1} + z^{n-2}+ \cdots + z + 1 = (z-Z)(z-Z^2)\cdots(z-Z^{n-1})$$ Let $z=1 \implies n = (1-Z)(1-Z^2)\cdots(1-Z^{n-1})$. Take modulus on both sides: $$n = |1-Z|\cdot|1-Z^2|\cdots|1-Z^{n-1}|$$ On the other hand: $|1-Z^k|=|1-\cos{\frac{2k\pi}{n}} - i\sin{\frac{2k\pi}{n}}|=2\sin{\frac{k\pi}{n}}$. Therefore $n=\displaystyle\prod_{k=1}^{n-1}(2\sin{\frac{k\pi}{n}})=2^{n-1}\displaystyle\prod_{k=1}^{n-1}(\sin{\frac{k\pi}{n}})$ or $\displaystyle\prod_{k=1}^{n-1}(\sin{\frac{k\pi}{n}})=\frac{n}{2^{n-1}}$.