TITLE: On determinants formed by binomial coefficients
QUESTION [23 upvotes]: 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, $(q+4)(q-1)/2 =7$.

REPLY [13 votes]: For the specific determinant in the question there is a slightly more direct proof than Lemma 3 in Krattenthaler. We can evaluate it using the Jacobi-Trudi identity. Let's denote  $\lambda _i=(q-i+1)(q-1)$. The determinant in question, $\det\left(\binom{q^2-1}{qi-q+j}\right)$, is equal (up to sign) to the determinant $\det\left(e_{\lambda_i-i+j}\right)$ when you put all variables equal to $1$. But this is equal to the Schur polynomial of the conjugate partition $\lambda^{\star}=(q^{q-1}(q-1)^{q-1}\cdots 1^{q-1}0^{q-1})$ evaluated at $1,1,\dots,1$. (I added a bunch of zeros to the partition so that the number of parts matches the number of variables.)
There is a simple product formula for Schur polynomials evaluated at $1,1,\dots,1$, and in this case it gives
$$s_{\lambda^{\star}}(1,1,\dots,1)=\prod_{i<j}\frac{\lambda^{\star}_i-\lambda^{\star}_j+j-i}{j-i}.$$
The denominator has the form $1!2!\cdots (q^2-2)!$ so the highest power of $q$ dividing it is $q^{\frac{(q-1)(q^2-2)}{2}}$. For the numerator, notice that $\lambda^{\star}_i-\lambda^{\star}_j+j-i$ is divisible by $q$ iff $j-i$ is divisible by $q-1$. This means that exactly $(q-1)\binom{q+1}{2}$ factors in the numerator are divisible by $q$. Out of these the pairs $(i,j)=(r,q^2-q+r)$ for $r\in\{1,2,\dots,q-1\}$ are divisible by $q^2$. So the power of $q$ dividing your determinant is exactly
$$(q-1)\binom{q+1}{2}+(q-1)-\frac{(q-1)(q^2-2)}{2}=\frac{(q-1)(q+4)}{2}$$
as desired.