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, $(q+4)(q-1)/2 =7$.