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

If you want a proof without induction:
Let $\displaystyle P(X)=\sum\limits_{i=0}^{n-1} X^i$. Notice that the roots of $P$ are the $\displaystyle e^{i2k\pi/n}$ for $1 \leq k \leq n-1$, hence $\displaystyle P(X)= \prod\limits_{i=1}^{n-1} \left(X-e^{i 2k\pi/n} \right)$. Therefore, $$n=P(1)= \prod\limits_{i=1}^n \left( 1- e^{i2k\pi/n} \right)= \prod\limits_{i=1}^n e^{ik\pi/n} \left( e^{-ik\pi/n}-e^{ik\pi/n} \right) =2^{n-1} \prod\limits_{i=1}^{n-1} \sin \left( \frac{k\pi}{n} \right)$$
