Calculating $\prod (\omega^j - \omega^k)$ where $\omega^n=1$. Let $1, \omega, \dots, \omega^{n-1}$ be the roots of the equation $z^n-1=0$, so that the roots form a regular $n$-gon in the complex plane. I would like to calculate
$$ \prod_{j \ne k} (\omega^j - \omega^k)$$
where the product runs over all $j \ne k$ with $0 \le j,k < n$.

My attempt so far
Noting that if $k-j = d$ then $\omega^j - \omega^k = \omega^j(1-\omega^d)$, I can re-write the product as
$$ \prod_{d=1}^{\lfloor n/2 \rfloor} \omega^{n(n-1)/2}(1-\omega^d)^n$$
I thought this would be useful but it hasn't led me anywhere.
Alternatively I could exploit the symmetry $\overline{1-\omega^d} = 1-\omega^{n-d}$ somehow, so that the terms in the product are of the form $|1-\omega^d|^2$. I tried this and ended up with a product which looked like
$$\prod_{j=0}^{n-1} |1 - \omega^j|^n $$
(with awkward multiplicative powers of $-1$ left out). This appears to be useful, but calculating it explicitly is proving harder than I'd have thought.
The answer I'm expecting to find is something like $n^n$.

My motivation for this comes from Galois theory. I'm trying to calculate the discriminant of the polynomial $X^n+pX+q$. I know that it must be of the form $ap^n+bq^{n-1}$ for some $a,b \in \mathbb{Z}$, and putting $p=0,q=-1$, the polynomial becomes $X^n-1$. This has roots $1, \omega, \dots, \omega^{n-1}$, so that $(-1)^{n-1}b$ is (a multiple of) the product you see above. An expression for $a$ can be found similarly by setting $p=-1,q=0$.

First, note that 
$$\prod_{k=1}^{n-1} (1-w^k) = n$$
The proof is that $\prod_{k=1}^{n-1}(x-w^k) = 1+x+x^2+...+x^{n-1}$, then substitute $x=1$.
Now, you can rewrite:
$$\prod_{j\neq k} (w^j-w^k) = \prod_{j=0}^{n-1} \prod_{i=1}^{n-1} (w^j-w^{i+j})$$
$$= \prod_{j=0}^{n-1} w^{j(n-1)} n = n^n w^{\frac{n(n-1)^2}2}$$
If $n$ is odd, then $w^{\frac{n(n-1)^2}2}= 1$, otherwise $w^{\frac{n(n-1)^2}2}=-1$.  So we can write our formula as $(-1)^{n-1}n^n$ or as $-(-n)^n$.
