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 $n \in N$?
 A: 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 $i^{n-1} = e^{\frac{i(n-1)\pi}{2}}$
A: 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
$$ \lim_{z\to 0}f(z) = -\frac{2}{2^n}.\tag{2}$$
A: Using De Moivre's formula for odd $n=2m+1$,
and writing $\cos x=c,\sin x=s$
$$i\sin(2m+1)x=(i s)^{2m+1}+\binom{2m+1}2(i s)^{2m-1}c^2+\binom{2m+1}4(i s)^{2m-3}c^4+\cdots+\binom{2m+1}{2m}(is)c^{2m}$$
$$=i^{2m+1}[s^{2m+1}-\binom{2m+1}2s^{2m-1}(1-s^2)+\binom{2m+1}4s^{2m-3}(1-s^2)^2+\cdots+\binom{2m+1}{2m}(-1)^ms(1-s^2)^m]$$
$$\iff s^{2m+1}\left[1+\binom{2m+1}2+\binom{2m+1}4+\cdots+\binom{2m+1}{2m}\right]+\cdots= (-1)^m\sin(2m+1)x$$
$$\iff s^{2m+1}\left[1+1\right]^{2m+1-1}- (-1)^m\sin(2m+1)x=0$$
Now if $\sin(2m+1)x=\sin(2m+1)a,$
$(2m+1)x=2r\pi+(2m+1)a, x=a+\dfrac{2r\pi}{2m+1}$ where $-m\le r\le m$
$\implies2^{2m}\prod_{r=-m}^m\sin\left(a+\dfrac{2r\pi}{2m+1}\right)=-(-1)^m\sin(2m+1)a$
Now for $-m\le r<0, r=-t$(say) $\implies m\ge t>0$
$\sin\left(a+\dfrac{2r\pi}{2m+1}\right)=\sin\left(a-\dfrac{2t\pi}{2m+1}\right)$
$=-\sin\left(a-\dfrac{2t\pi}{2m+1}+\pi\right)=-\sin\left(a+\dfrac{(2m+1-2t)\pi}{2m+1}\right)$
As $m\ge t>0,-2m\le -2t<0\iff1\le2m+1-2t<2m+1$
$\implies2^{2m}\prod_{t=0}^{2m}\sin\left(a+\dfrac{t\pi}{2m+1}\right)(-1)^m=-(-1)^m\sin(2m+1)a$
Similarly for the even $n=2m$(say)
