Let $\eta=e^{\frac{2\pi i}n}$, an $n$-th root of unity. For pedagogical reasons and inspiration, I ask to see different proofs (be it elementary, sophisticated, theoretical, etc) for the following product evaluation.



> 
> If $T(n)=\frac{(3n-2)(n-1)}2$ and $i=\sqrt{-1}$ then
>  $$\prod\_{j<k}^{0,n-1}(\eta^k-\eta^j)=n^{\frac{n}2}i^{T(n)}.$$
> 
> 
> 



Different derivations of the value of $\prod_{0\leq j<k<n}(\eta^k-\eta^j)$
We first find the norm; we then determine the argument.


Call the product you wrote $A\_n$. Then $A\_n^2 = \prod\_{j<k}^{0,n-1} (\eta^k - \eta^j)^2 = Disc(x^n - 1) = (-1)^{\frac{n (n -1)}{2}}Res(x^n - 1, n x^{n - 1})$


$= (-1)^{\frac{n(n-1)}{2}} n^n \prod\_{0 \leq i < n, 0 \leq j < n-1} (\eta^i - 0)$


All terms in the expression except $n^n$ have norm $1$, so we have that $|A\_n| = n^{\frac{n}{2}}$. We therefore only need to figure out the argument of $A\_n$.


Let $\eta' = e^\frac{2 \pi i}{2n}$ be the square root of $\eta$. We can rewrite $A\_n = \prod\_{0\leq j<k<n} \eta'^{k + j} (\eta'^{k - j} - \eta'^{j - k})$. Note that the second term is a difference of (unequal) conjugates where the minuend has positive imaginary part (and the subtrahend therefore negative imaginary part), and therefore will always have argument $\frac{\pi}{2}$. So let us concentrate on the argument of the first term, $\prod\_{0 \leq j < k < n} \eta'^{k +j}$. We can do this by finding $\sum\_{0 \leq j < k < n} j + k$. 


$\sum\_{0 \leq j < k < n} j + k = \left(\sum\_{0 \leq j < k < n} j\right) + \left(\sum\_{0 \leq j < k < n} k\right)$


$= \left(\sum\_{0 \leq j <n} (n - j - 1)j\right) + \left(\sum\_{0\leq k<n} k\*k\right)$


$= \sum\_{0 \leq j < n} (n - j - 1)j + j\*j = \sum\_{0 \leq j < n} (n - 1)j$


$= (n - 1) \frac{n (n - 1)}{2}$


We therefore end up with an argument of $\frac{n(n - 1)}{2} \frac{\pi}{2} + \frac{n (n - 1)^2}{2} \frac{2 \pi}{2n} = \frac{(3n^2 - 5n + 2)\pi}{4}$. We finally have that:


The norm of $A\_n$ is $n^\frac{n}{2}$, and the argument is $\frac{(3n^2 - 5n + 2)\pi}{4}$. Correspondingly, we have that $A\_n = n^{\frac{n}{2}} i^{T(n)}$, as desired.


