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)
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n}=\frac{n}{2^{n-1}}~(n\geq 2)$$

Here is a non inductive proof:
$LHS=\prod(\frac{e^{ik\pi/n}-e^{ik\pi/n}}{2i})$
$=(\frac{1}{2i})^{n-1}\prod e^{ik\pi/n}\prod(1-e^{i2k\pi/n})$
$=(\frac{1}{2i})^{n-1}e^{\frac{i\pi}{n}(\frac{n(n-1)}{2})}\prod(1-e^{i2k\pi/n})$
$=\frac{1}{2^{n-1}}\prod(1-e^{i2k\pi/n})$. 
Now $\prod(1-e^{i2k\pi/n})=n$ can be proved as follows: Note that each $r_k:=e^{i2k\pi/n}$ is a $n$th root of unity which is not 1. Also the roots of $x^{n-1}+\cdots x+1$ are precisely these numbers $r_k$. So $(x-r_1)\cdots(x-r_{n-1})\equiv x^{n-1}+\cdots x+1$. Letting $x=1$ we get $\prod(1-e^{i2k\pi/n})=n$ which proves the result.
