How can i prove this identity (by mathematical induction) (rational product of sines) I would appreciate if somebody could help me with the following problem: 
Q: proof? (by mathematical induction)
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n}=\frac{n}{2^{n-1}}~(n\geq 2)$$

Let
$$S_n=\prod_{k=1}^{n-1}\sin \frac{k\pi}{n}.$$
We solve the equation $(z+1)^n=1$ for $z\in\mathbb{C}$ we find
$$z=e^{i2k\pi/n}-1=2ie^{ik\pi/n}\sin\frac{k\pi}{n}=z_k,\quad 0,\ldots,n-1.$$
Moreover $(x+1)^n-1=x\left((x+1)^{n-1}+(x+1)^{n-2}+\cdots+(x+1)+1\right)=xP(x).$
The roots of $P$ are $z_k, k=1,\ldots ,n-1$. By the relation between polynomial's cofficients of $P$ and its roots we have $$\sigma_{n-1}=(-1)^{n-1}n=\prod_{k=1}^{n-1}z_k.$$
In another way, we have
$$\prod_{k=1}^{n-1}z_k=2^{n-1}i^{n-1}\left(\prod_{k=1}^{n-1}e^{ik\pi/n}\right)\left(\prod_{k=1}^{n-1}\sin \frac{k\pi}{n}\right)=2^{n-1}i^{n-1}e^{i\pi(1+2+\cdots+(n-1)/n)}S_n=2^{n-1}(-1)^{n-1}S_n.$$
We conclude.
