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 $n \in N$?

This lemma is often used in the proof of the multiplication formula for the $\Gamma$ function. I think it is worth mentioning that both this lemma and the multiplication formula can be proved along the same lines. Since:
$$\sin x = x \prod_{n=1}^{+\infty}\left(1-\frac{x^2}{\pi^2 n^2}\right)$$
we have that the meromorphic function defined by:
$$ f(z) = \frac{\prod_{k=1}^{n}\sin\left(z+\frac{\pi k}{n}\right)}{\sin(nz)}\tag{1}$$ 
has no zeroes and no poles, since $z$ is a simple zero for $\sin(nz)$ iff it is a simple zero for $\prod_{k=1}^{n}\sin\left(z+\frac{\pi k}{n}\right)$. So we have that $f(z)$ is a non-vanishing entire function. It is not difficult to check that $f(z)$ is an order-$0$ entire function, then to prove that $f(z)$ is bounded on $\mathbb{C}$. 
Liouville's theorem hence gives that $f(z)$ is constant. The last step is just to prove that
$$ \lim_{z\to 0}f(z) = -\frac{2}{2^n}.\tag{2}$$
