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 with positive 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.

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.