How prove $\sin \left( \alpha+\frac{\pi }{n} \right) \cdots \sin \left( \alpha+\frac{n\pi }{n} \right) =-\frac{\sin n\alpha}{2^{n-1}}$? How prove 
$$\prod_{k=1}^{n}\sin \left( \alpha+\frac{\pi k }{n}\right) =-\frac{\sin n\alpha}{2^{n-1}}$$ 
for $n \in N$?

Since, $\sin \theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}$, 
Hence, $$\begin{align}\prod_{j=1}^{n-1}\sin \left(\alpha+\frac{j\pi}{n}\right)&=\prod_{j=1}^{n-1}\left(\frac{e^{i(\alpha+j\pi/n)}-e^{-i(\alpha+j\pi/n)}}{2i}\right) \\&= \left(\frac{-1}{2i}\right)^{n-1}e^{-((n-1)\alpha + i\frac{(n-1)\pi}{2})}\prod_{j=1}^{n-1} \left(1-e^{2i\alpha+2ij\pi/n}\right) \tag{1} \\&= \left(\frac{-1}{2i}\right)^{n-1}e^{-((n-1)\alpha + i\frac{(n-1)\pi}{2})}\sum_{j=0}^{n-1} e^{2ij\alpha} \tag{2} \\ &= \left(\frac{-1}{2i}\right)^{n-1}e^{-((n-1)\alpha + i\frac{(n-1)\pi}{2})}.\frac{e^{2in\alpha}-1}{e^{2i\alpha}-1} \tag{3}\\&= \left(\frac{-1}{2i}\right)^{n-1}e^{-i\frac{(n-1)\pi}{2}} \frac{\sin n\alpha}{\sin \alpha} \tag{4} \\ &= \frac{\sin n\alpha}{2^{n-1}\sin \alpha}\end{align}$$
Explanations:
$(1)$ Factored out $\displaystyle \prod\limits_{j=1}^{n-1}e^{-i(\alpha+j\pi/n)}$ from the product. $$\displaystyle \prod\limits_{j=1}^{n-1}e^{-i(\alpha+j\pi/n)} = \exp\left(\frac{i\pi}{n}\sum\limits_{j=1}^{n-1} j\right) = e^{\frac{i(n-1)\pi}{2}}$$
$(2)$ Used the fact that $\displaystyle 1+z+\cdots + z^{n-1} = \prod\limits_{j=1}^{n-1} (1-z\omega^j)$, where, $\omega = e^{2i\pi/n}$ is the $n^{th}$ root of unity. Here, $z = e^{2i\alpha}$
$(3)$ $\displaystyle 1+z+\cdots + z^{n-1} = \frac{z^n-1}{z-1}$
$(4)$ $\displaystyle \frac{e^{2in\alpha}-1}{e^{2i\alpha}-1} = e^{i(n-1)\alpha}\frac{e^{in\alpha} - e^{-in\alpha}}{e^{i\alpha} - e^{i\alpha}} = e^{i(n-1)\alpha}\frac{\sin n\alpha}{\sin \alpha}$ and $i^{n-1} = e^{\frac{i(n-1)\pi}{2}}$
