I was working on a complex analysis problem in "Berkeley Problems in Mathematics", which was asking me to prove that some product is equal to $n$. I had reduced the problem to proving a trigonometric identity, but I couldn't prove it although I spent much time on it. Then, I checked the solution, and it solves the problem with a quite different approach that doesn't even enter to trigonometric expressions at all. But, still, this nice identity that I hadn't encountered before must be true then! I was wondering if people can think of a direct proof for that?
Yes, here is the identity: For $n\geq 2$ integer, we have
$$2^{n-1} \sin \left(\frac{\pi}{n}\right)\sin \left(\frac{2\pi}{n}\right)\cdots \sin \left((n-1)\frac{\pi}{n}\right)=n$$
Thanks!
