TITLE: Determinants question with factorials
QUESTION [6 upvotes]: $$A= \begin{pmatrix}
     n! & (n+1)! & (n+2)!\\
     (n+1)! & (n+2)! & (n+3)! \\
     (n+2)! & (n+3)! & (n+4)!
     \end{pmatrix}$$
And $D=\det(A)$,
We have to prove that $D/(n!)^3 - 4$ is divisible by $n$.
I did it by simplifying the determinant using some row and column operations and finally ended up in getting a cubic polynomial in $n$. Then I subtracted 4 from the polynomial and got the result.
Is there any other way I could prove the result quicker and in a better manner?

REPLY [15 votes]: Following hypergeometric's answer, we have an exact expression for $D/(n!)^3$.  All that remains is to show $D/(n!)^3 \equiv 4 \pmod{n}$.  Since we just want the determinant of $B$ mod $n$ we can reduce the matrix entries mod $n$:
$$\begin{bmatrix}
1&1&2\\
1&2&6\\
2&6&24
\end{bmatrix},$$
and a routine calculation gives a determinant $4$.