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Changed arXiv link to point to abstract
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edit approved Sep 25, 2015 at 0:06
LSpice
edit approved Sep 25, 2015 at 0:06
LSpice
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Looking at Krattenthaler's famous "determinant calculus" surveyKrattenthaler's famous "determinant calculus" survey (also, suggested by Per Alexandersson) we find Theorem 26 in there, which answers a more general question, and should (most likely) yield the claim in the OP.

Theorem Let $q$ be a nonnegative integer, and let $L_1,\ldots,L_q$ and $A, B$ be indeterminates. Then, \begin{equation*} \det\left( \binom{BL_i+A}{L_i+j}\right) = \frac{\prod_{i < j}(L_i-L_j)}{\prod_{i=1}^q(L_i+q)!}\prod_{i=1}^q\frac{(BL_i+A)!}{((B-1)L_i+A-1)!}\prod_{i=1}^q(A-Bi+1)_{i-1}, \end{equation*} where $(a)_k$ is the usual Pochhammer symbol.

The matrix in the OP is given by \begin{equation*} M_{ij} = \binom{q^2-1}{q(i-1)+j-1}, \end{equation*} thus we can select $A=q^2-1$, $B=0$, $L_i = q(i-1)-1$, and apply the above result, and hopefully things fall into place.

Looking at Krattenthaler's famous "determinant calculus" survey (also, suggested by Per Alexandersson) we find Theorem 26 in there, which answers a more general question, and should (most likely) yield the claim in the OP.

Theorem Let $q$ be a nonnegative integer, and let $L_1,\ldots,L_q$ and $A, B$ be indeterminates. Then, \begin{equation*} \det\left( \binom{BL_i+A}{L_i+j}\right) = \frac{\prod_{i < j}(L_i-L_j)}{\prod_{i=1}^q(L_i+q)!}\prod_{i=1}^q\frac{(BL_i+A)!}{((B-1)L_i+A-1)!}\prod_{i=1}^q(A-Bi+1)_{i-1}, \end{equation*} where $(a)_k$ is the usual Pochhammer symbol.

The matrix in the OP is given by \begin{equation*} M_{ij} = \binom{q^2-1}{q(i-1)+j-1}, \end{equation*} thus we can select $A=q^2-1$, $B=0$, $L_i = q(i-1)-1$, and apply the above result, and hopefully things fall into place.

Looking at Krattenthaler's famous "determinant calculus" survey (also, suggested by Per Alexandersson) we find Theorem 26 in there, which answers a more general question, and should (most likely) yield the claim in the OP.

Theorem Let $q$ be a nonnegative integer, and let $L_1,\ldots,L_q$ and $A, B$ be indeterminates. Then, \begin{equation*} \det\left( \binom{BL_i+A}{L_i+j}\right) = \frac{\prod_{i < j}(L_i-L_j)}{\prod_{i=1}^q(L_i+q)!}\prod_{i=1}^q\frac{(BL_i+A)!}{((B-1)L_i+A-1)!}\prod_{i=1}^q(A-Bi+1)_{i-1}, \end{equation*} where $(a)_k$ is the usual Pochhammer symbol.

The matrix in the OP is given by \begin{equation*} M_{ij} = \binom{q^2-1}{q(i-1)+j-1}, \end{equation*} thus we can select $A=q^2-1$, $B=0$, $L_i = q(i-1)-1$, and apply the above result, and hopefully things fall into place.

Fixed missing parenthesis
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Julian Rosen
Julian Rosen
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Looking at Krattenthaler's famous "determinant calculus" survey (also, suggested by Per Alexandersson) we find Theorem 26 in there, which answers a more general question, and should (most likely) yield the claim in the OP.

Theorem Let $q$ be a nonnegative integer, and let $L_1,\ldots,L_q$ and $A, B$ be indeterminates. Then, \begin{equation*} \det\left( \binom{BL_i+A}{L_i+j}\right) = \frac{\prod_{i < j}(L_i-L_j)}{\prod_{i=1}^q(L_i+q)!}\prod_{i=1}^q\frac{(BL_i+A)!}{(B-1)L_i+A-1)!}\prod_{i=1}^q(A-Bi+1)_{i-1}, \end{equation*}\begin{equation*} \det\left( \binom{BL_i+A}{L_i+j}\right) = \frac{\prod_{i < j}(L_i-L_j)}{\prod_{i=1}^q(L_i+q)!}\prod_{i=1}^q\frac{(BL_i+A)!}{((B-1)L_i+A-1)!}\prod_{i=1}^q(A-Bi+1)_{i-1}, \end{equation*} where $(a)_k$ is the usual Pochhammer symbol.

The matrix in the OP is given by \begin{equation*} M_{ij} = \binom{q^2-1}{q(i-1)+j-1}, \end{equation*} thus we can select $A=q^2-1$, $B=0$, $L_i = q(i-1)-1$, and apply the above result, and hopefully things fall into place.

Looking at Krattenthaler's famous "determinant calculus" survey (also, suggested by Per Alexandersson) we find Theorem 26 in there, which answers a more general question, and should (most likely) yield the claim in the OP.

Theorem Let $q$ be a nonnegative integer, and let $L_1,\ldots,L_q$ and $A, B$ be indeterminates. Then, \begin{equation*} \det\left( \binom{BL_i+A}{L_i+j}\right) = \frac{\prod_{i < j}(L_i-L_j)}{\prod_{i=1}^q(L_i+q)!}\prod_{i=1}^q\frac{(BL_i+A)!}{(B-1)L_i+A-1)!}\prod_{i=1}^q(A-Bi+1)_{i-1}, \end{equation*} where $(a)_k$ is the usual Pochhammer symbol.

The matrix in the OP is given by \begin{equation*} M_{ij} = \binom{q^2-1}{q(i-1)+j-1}, \end{equation*} thus we can select $A=q^2-1$, $B=0$, $L_i = q(i-1)-1$, and apply the above result, and hopefully things fall into place.

Looking at Krattenthaler's famous "determinant calculus" survey (also, suggested by Per Alexandersson) we find Theorem 26 in there, which answers a more general question, and should (most likely) yield the claim in the OP.

Theorem Let $q$ be a nonnegative integer, and let $L_1,\ldots,L_q$ and $A, B$ be indeterminates. Then, \begin{equation*} \det\left( \binom{BL_i+A}{L_i+j}\right) = \frac{\prod_{i < j}(L_i-L_j)}{\prod_{i=1}^q(L_i+q)!}\prod_{i=1}^q\frac{(BL_i+A)!}{((B-1)L_i+A-1)!}\prod_{i=1}^q(A-Bi+1)_{i-1}, \end{equation*} where $(a)_k$ is the usual Pochhammer symbol.

The matrix in the OP is given by \begin{equation*} M_{ij} = \binom{q^2-1}{q(i-1)+j-1}, \end{equation*} thus we can select $A=q^2-1$, $B=0$, $L_i = q(i-1)-1$, and apply the above result, and hopefully things fall into place.

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Suvrit
Suvrit
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Looking at Krattenthaler's famous "determinant calculus" survey (also, suggested by Per Alexandersson) we find Theorem 26 in there, which answers a more general question, and should (most likely) yield the claim in the OP.

Theorem Let $q$ be a nonnegative integer, and let $L_1,\ldots,L_q$ and $A, B$ be indeterminates. Then, \begin{equation*} \det\left( \binom{BL_i+A}{L_i+j}\right) = \frac{\prod_{i < j}(L_i-L_j)}{\prod_{i=1}^q(L_i+q)!}\prod_{i=1}^q\frac{(BL_i+A)!}{(B-1)L_i+A-1)!}\prod_{i=1}^q(A-Bi+1)_{i-1}, \end{equation*} where $(a)_k$ is the usual Pochhammer symbol.

The matrix in the OP is given by \begin{equation*} M_{ij} = \binom{q^2-1}{q(i-1)+j-1}, \end{equation*} thus we can select $A=q^2-1$, $B=0$, $L_i = q(i-1)-1$, and apply the above result, and hopefully things fall into place.