Proving $\prod_k \sin \pi k / n = n / 2^{n-1}$ I am stuck trying to prove 
$$\prod_{k=1}^{n-1} \sin {\pi k \over n} = {n \over 2^{n-1}}$$
and I'd appreciate help. 
What I have done so far:
$z^n - 1 = \prod_{k=1}^n (z - \xi^k)$ where $\xi = e^{2 \pi i k\over n}$. Dividing both sides by $z-1$ we get
$$ {z^n - 1 \over z - 1} = \prod_{k=1}^{n-1} (z - \xi^k)$$
Taking the limit $z \to 1$ on both sides,
$$ n = \prod_{k=1}^{n-1} (1 - \xi^k)$$
So I'm getting close to what I want to prove but unfortunately, this is where I am stuck! 

How to proceed form here?

 A: You might try using the identity $$\sin\left(\frac{\pi k}{n}\right)=\frac{e^{i\pi k/n}-e^{-i\pi k/n}}{2i}= -\frac{e^{-\pi i k/n}}{2i}\left(1-e^{2i\pi k/n}\right)$$ 
to convert the desired product to something that resembles the computation you have done so far.
A: Let
$$
P=\prod_{k=1}^{n-1}\sin\frac{k\pi}{n},\quad a:=e^{\frac{\pi i}{n}}=\cos\frac{\pi}{n}+i\sin\frac{\pi}{n},\quad k=0,1,\ldots,n-1.
$$
Notice that the roots of $z^n-1$ are:
$$
a^0=1,\, a^2,\,a^4,\, \ldots,\, a^{2(n-1)},
$$
therefore
we have
$$
\sum_{l=0}^{n-1}z^l=\frac{z^n-1}{z-1}=\prod_{k=1}^{n-1}(z-a^{2k}) \quad \forall z\ne 1.
$$
By letting $z\to1$ we get:
$$\tag{1}
n=\prod_{k=1}^{n-1}(1-a^{2k})=\prod_{k=1}^{n-1}a^k(a^{-k}-a^k)=\left(\prod_{k=1}^{n-1}a^k\right)\cdot\left(\prod_{k=1}^{n-1}(a^{-k}-a^k)\right)
$$
But
$$\tag{2}
\prod_{k=1}^{n-1}(a^{-k}-a^k)=\prod_{k=1}^{n-1}\left[-2i\sin\frac{k\pi}{n}\right]=P\prod_{k=1}^{n-1}(-2i)=(-2i)^{n-1}P,
$$
and
$$\tag{3}
\prod_{k=1}^{n-1}a^k=a^{1+2+\ldots+(n-1)}=a^{n(n-1)/2}=e^{i(n-1)\pi/2}=i^{n-1}
$$
Combining (10, (2), and (3), we get:
$$
n=i^{n-1}(-2i)^{n-1}P=(-2)^{n-1}i^{2(n-1)}P=2^{n-1}P.
$$
Hence
$$
P=\frac{n}{2^{n-1}}.
$$
