Representing $ 2 i \sin(2 \pi n z) $ as a product I'm currently reading a surprising proof of the Quadratic Reciprocity Law, which uses the following function:
$ f(z) = 2 i \sin(2 \pi z) $
One of its properties is that 
$$\frac{f(nz)}{f(z)} = \prod \limits_{k=1}^{\frac{n-1}{2}} f\left(z + \frac{k}{n}\right)f\left(z - \frac{k}{n}\right)$$
and the step that I can't get over is the following:
We notice that $$ f(nz) = 2 i \sin(2 \pi n z) = \exp(2 i \pi n z) - \exp(-2 i \pi n z) $$ and it equals $$\prod\limits_{k=1}(\zeta^ka - \zeta^{-k}b)$$
where $ \zeta = \exp(2 i \pi /n), a = \exp(2 \pi z), b = \exp(-2 \pi z)$
I don't know where dthis comes from. I only know one formula for $ a^n - b^n $ and it has nothing to do with such form. I would appreciate some clarification
 A: It's important to mention that $n$ is odd! So, $z^n-1$ polynomial has $n$ complex roots:
$$z^n-1=\prod_{k=0}^{n-1}\left ( z-e^{2\pi i \frac{k}{n} } \right )=\prod_{k=0}^{n-1}\left ( z-\zeta^k \right )$$
Now $\left \{ -2k \pmod{n} \mid k=0..n-1 \right \} = \left \{ 0,1,2,3,...,n-1 \right \}, \forall n \geq 3$, $n$-odd. Otherwise, if we assume $-2i \equiv -2j \pmod{n}$, for $i\neq j$, or even better $0\leq i<j \leq n-1$, then $2(j-i) \equiv 0 \pmod{n}$ or $0<2(j-i)=qn<2n$ which means either $q=0$ (bad) or $q=1$ (bad too because $n$ is odd).
Also $\zeta^k=\zeta^{k+n}$, i.e. periodic by $n$.
As a result, the expression above can be written as (with re-arrangements):
$$z^n-1=\prod_{k=0}^{n-1}\left ( z-\zeta^{-2k} \right )$$
Which is $$\prod_{k=0}^{n-1}\left ( z-\zeta^{-2k} \right )=\prod_{k=0}^{n-1} \frac{1}{\zeta^{k}} \cdot \prod_{k=0}^{n-1}\left ( z\zeta^{k}-\zeta^{-k} \right )$$
And $$\prod_{k=0}^{n-1} \frac{1}{\zeta^{k}}=\prod_{k=1}^{n-1} \frac{1}{\zeta^{k}}=\frac{1}{\zeta^{1+2+..+n-1}}=\frac{1}{\zeta^{\frac{n(n-1)}{2}}}=1$$
$n$-odd! And this is:
$$z^n-1= \prod_{k=0}^{n-1}\left ( z\zeta^{k}-\zeta^{-k} \right )$$
Finally, let's substitute $z=\frac{a}{b}$ and we have
$$\left ( \frac{a}{b} \right )^n-1=\prod_{k=0}^{n-1}\left (\frac{a}{b} \cdot \zeta^{k}-\zeta^{-k} \right )$$
which is:
$$\frac{a^n-b^n}{b^n}=\frac{1}{b^n} \prod_{k=0}^{n-1}\left (a \zeta^{k}-b\zeta^{-k} \right )$$
Or
$$a^n-b^n=\prod_{k=0}^{n-1}\left (a \zeta^{k}-b\zeta^{-k} \right )$$
I was partially inspired by this.
