Let $q$ be a number. Let us consider the $q^2-1$-th line of the Pascal triangle (i.e. numbers ${{q^2-1} \choose i}$, $i=0,1,...q^2-1$). We have $q^2$ numbers. 


Let us form naively a $q \times q$ matrix from them. Example for $q=2$:


\[\begin{pmatrix} 1 & 3 \\
3 & 1 \end{pmatrix}\]


And let us take its determinant: $-8$ if $q=2$. 


How to prove that if $q$ is prime then $q$ enters in this determinant with 
$(q+4)(q-1)/2 $-th exponent?


Example: $q=2$, $(q+4)(q-1)/2 =3$. Really, $8=2^3$. 


Case $q=3$: the binomial coefficients are $1,8,28,56,70,56,28,8,1$.


The matrix: 


\[\begin{pmatrix} 1 & 8 & 28 \\
56 & 70 & 56 \\
28 & 8 & 1 \end{pmatrix}\]


The determinant is $2 \cdot 3^7 \cdot 7$. Really, $(q+4)(q-1)/2 =7$. 



On determinants formed by binomial coefficients
Looking at [Krattenthaler's famous "determinant calculus" survey](http://arxiv.org/abs/math/9902004) (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](https://en.wikipedia.org/wiki/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.


