It is given that: $$\sin\frac{\pi }{n} \sin\frac{2\pi }{n}\cdots\sin\frac{(n-1)\pi }{n}=\frac{n}{2^{n-1}}\tag1$$It is asked to use the above identity to evaluate the following improper integral:$$\int_0^\pi \log(\sin x) \mathrm dx$$
I used the definition of the integral in terms of Riemann sums:$$\begin{align*}\int_0^\pi \log(\sin x) \mathrm dx&=\lim_{n\rightarrow \infty }\frac{\pi }{n}\left[\sum_{k=1}^{k=n-1}\log\left(\sin\left(\frac{k\pi }{n}\right)\right)\right]\\&=\lim_{n\rightarrow \infty }\frac{\pi }{n}\left[\log\left(\sin\frac{\pi }{n}\sin\frac{2\pi }{n}\cdots\sin\frac{(n-1)\pi }{n}\right)\right]\\&=\lim_{n\rightarrow \infty }\frac{\pi }{n}\Big(\log n-(n-1)\log 2\Big)\\&=-\pi \log 2\end{align*}$$
However, this integral is improper, so $\log(\sin(\pi ))=\log(0)=-\infty $. I am kind of cheating in my solution, because the Riemann sum above should be: $$\sum_{k=1}^{k=n-1}\log\left(\sin\left(\frac{k\pi }{n}\right)\right)+\frac{\pi }{n}\log\left(\sin\left(\frac{n\pi }{n}\right)\right)\;$$ but I have no idea how to deal with the last term of the sum since $\sin\left(\frac{n\pi }{n}\right)=\sin(\pi)=0 $.
Can anyone show me how to deal with this?Also, how to prove the identity $(1)$?