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Let $eta=e^frac2pi in$, 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
$$prod_j<k^0,n-1(eta^k-eta^j)=n^fracn2i^T(n).$$

Your are asking about determinant of the Schur Matrix. So you can use original Schur's article or another classical expositions mentioned at Mathworld.

We first find the norm; we then determine the argument.

Call the product you wrote $A_n$. Then $A_n^2 = prod_j<k^0,n-1 (eta^k - eta^j)^2 = Disc(x^n - 1) = (-1)^fracn (n -1)2Res(x^n - 1, n x^n - 1)$

$= (-1)^fracn(n-1)2 n^n prod_0 leq i < n, 0 leq j < n-1 (eta^i - 0)$

All terms in the expression except $n^n$ have norm $1$, so we have that $|A_n| = n^fracn2$. We therefore only need to figure out the argument of $A_n$.

Let $eta' = e^frac2 pi i2n$ be the square root of $eta$. We can rewrite $A_n = prod_0leq j<k<n eta'^k + j (eta'^k - j - eta'^j - k)$. 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 $fracpi2$. 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 $sum_0 leq j < k < n j + k$.

$sum_0 leq j < k < n j + k = left(sum_0 leq j < k < n jright) + left(sum_0 leq j < k < n kright)$

$= left(sum_0 leq j <n (n - j - 1)jright) + left(sum_0leq k<n k*kright)$

$= sum_0 leq j < n (n - j - 1)j + j*j = sum_0 leq j < n (n - 1)j$

$= (n - 1) fracn (n - 1)2$

We therefore end up with an argument of $fracn(n - 1)2 fracpi2 + fracn (n - 1)^22 frac2 pi2n = frac(3n^2 - 5n + 2)pi4$. We finally have that:

The norm of $A_n$ is $n^fracn2$, and the argument is $frac(3n^2 - 5n + 2)pi4$. Correspondingly, we have that $A_n = n^fracn2 i^T(n)$, as desired.

Your determinant is essentially the Van der Monde determinant $rm det(A),$ where $A$ is the $n times n$ matrix $[eta^(j-1)(k-1)].$

Note that $A$ is the character table of the cyclic group of order $n$ so that $Abar A^T= nI_n$, using the orthogonality relations for group characters, and $|rm det A| = n^fracn2.$ One can continue in this vein, but I will sketch a more general calculation of the determinant of the character table of a general finite group, which simplifies considerably in the case of cyclic groups.

Let $G$ be a finite group with $t$ conjugacy classes, say with representatives $g_1,g_2, ldots g_t.$ Let us label these classes so that $1_G = g_1, g_2,ldots ,g_s$ are exactly those class representatives which are conjugate to their inverses in $G$ and so that $g_s+2j = g_s+2j-1^-1$ for $1 leq j leq fract-s2.$

Let $chi_1,chi_2, ldots chi_t$ be the complex irreducible characters of $G.$

Let $B$ be the character table of $G$, which is the $t times t$ matrix $[chi_j(g_k)].$

By the orthogonality relations for group characters, we see that $bar B^TB$ is the diagonal matrix whose $j$-th diagonal entry is $|C_G(g_j)|.$

Let $pi in rm S_t$ be the permutation fixing $1,2,ldots,s$ and interchanges $s+2j-1$ and $s+2j$ for $1 leq j leq fract-s2.$ Let $P$ be the associated permutation matrix.$

Note that $BP = bar B$ since $BP$ has the same first $s$ columns as $B$ and has the columns corresponding to $g_s+2j-1$ and $g_s+2j = g_s+2j-1^-1$ interchanged for $1 leq fract-s2$ (note that the first $s$ columns of $B$ are real as $g_j$ is conjugate to $g_j^-1$ for $1 leq j leq s.$

Hence $rm detB^2 = (-1)^fract-s2 prod_j=1^t |C_G(g_j)|$ and
$rm detB = (i)^fract-s2 sqrt$ where $i = sqrt-1.$

Edit following Mark Wildon's comment: In fact, although it looks as though there is a free choice of $sqrt-1$ in the above, in the case that $G$ is cyclic of order $n$, if we define $eta$ as in the question as $eta = expfrac2 pi in,$ then we have to use the other square root of $-1$ in the above expression for $rm det(A),$ so
$rm detA = (-i)^fracn-s2 n^fracn2$ where $s =1 $ if $n$ is odd and $s = 2$ if $n$ is even.

Even later edit: Here are some remarks about the general character table determinant which can be seen directly without making a choice for $sqrt-1.$ Note that since $bar B = BP$ and $overlinerm detB = (-1)^fract-s2rm detB$, we have that $rm det(B) in mathbbR$ if and only if $t equiv s$ (mod $4$). When $t not equiv s$ (mod $4$), we see that $rm detB$ is pure imaginary. At present, I don't see a quick way (in the general case) to determine which choice of $sqrt-1$ to use in the earlier formula once a choice of $i$ is fixed for the character table entries.

Here is a proof using the logarithmic function $mathrmLi_1(z)=-log(1-z)$:

Let $P= displaystyleprod_substackj,k=0 \ j<k^n-1 (eta^k-eta^j)$. Take the logarithm:
beginalign*
log P & = sum_j<k log eta^k - sum_j<k mathrmLi_1(eta^j-k) \
& = sum_k=0^n-1 k cdot frac2pi i kn - sum_a=1^n-1 (n-a) mathrmLi_1(eta^-a) \
& = frac2pi in sum_k=0^n-1 k^2 - sum_a=1^n-1 a mathrmLi_1(eta^a).
endalign*
Call $S$ the second sum. We have
beginalign*
S & = sum_a=1^n-1 a mathrmLi_1(eta^a) = frac12 sum_a=1^n-1 bigl(a mathrmLi_1(eta^a)+(n-a)mathrmLi_1(eta^-a)bigr)\
& = fracn2 sum_a=1^n-1 mathrmLi_1(eta^a) + sum_a=1^n-1 a cdot bigl(mathrmLi_1(eta^a)-mathrmLi_1(eta^-a)bigr)
endalign*
The first sum is easy to compute and is equal to $-log n$. Regarding the second sum, the classical Fourier expansion for the Bernoulli polynomial $B_1(x)=x-frac12$ on $(0,1)$ gives
beginequation*
mathrmLi_1(eta^a)-mathrmLi_1(eta^-a) = 2pi i bigl(frac12 - fracanbigr).
endequation*
From there, it is not difficult to finish the computation
beginalign*
log P = fracn2 log n - fracpi i4 n(n-1) + frac3pi in sum_k=1^n-1 k^2 = fracn2 log n + fracpi i4 (n-1)(3n-2)
endalign*
which gives the desired value for $P$.