Determinants question with factorials $$A= \begin{pmatrix}
     n! & (n+1)! & (n+2)!\\
     (n+1)! & (n+2)! & (n+3)! \\
     (n+2)! & (n+3)! & (n+4)!
     \end{pmatrix}$$
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?
 A: Factorizing the matrix might help a bit.
$$A= \begin{pmatrix}
     n! & (n+1)! & (n+2)!\\
     (n+1)! & (n+2)! & (n+3)! \\
     (n+2)! & (n+3)! & (n+4)!
     \end{pmatrix}=n!\underbrace{\begin{pmatrix}
     1 & (n+1) &(n+1)^{\overline{2}}\\
     (n+1) & (n+1)^{\overline{2}} & (n+1)^{\overline{3}} \\
     \;\;(n+1)^{\overline{2}} & (n+1)^{\overline{3}} & (n+1)^{\overline{4}}
     \end{pmatrix}}_B$$
using the Pochhammer notation for a rising factorial. 
$D/(n!)^3$ is then just the determinant of $B$.
A: $$
\begin{align}
D&=\det\begin{bmatrix}
n!&(n+1)!&(n+2)!\\
(n+1)!&(n+2)!&(n+3)!\\
(n+2)!&(n+3)!&(n+4)!
\end{bmatrix}\\
&=n!(n+1)!(n+2)!
\det\small\begin{bmatrix}
1&1&1\\
n+1&n+2&n+3\\
(n+1)(n+2)&(n+2)(n+3)&(n+3)(n+4)\\
\end{bmatrix}\tag{1}\\
&=n!(n+1)!(n+2)!
\det\begin{bmatrix}
1&1&1\\
0&1&2\\
0&2n+4&4n+10\\
\end{bmatrix}\tag{2}\\
&=n!(n+1)!(n+2)!
\det\begin{bmatrix}
1&1&1\\
0&1&2\\
0&0&2\\
\end{bmatrix}\tag{3}\\[12pt]
&=2n!(n+1)!(n+2)!\tag{4}
\end{align}
$$
Explanation:
$(1)$: divide columns by $n!$, $(n+1)!$, $(n+2)!$ respectively
$(2)$: subtract $n+1$ times row $1$ from row $2$
$\phantom{\text{(2):}}$ subtract $(n+1)(n+2)$ times row $1$ from row $3$
$(3)$: subtract $2n+4$ times row $2$ from row $3$
$(4)$: evaluate the determinant
Therefore,
$$
\begin{align}
\frac{D}{n!^3}-4
&=2(n+1)^2(n+2)-4\\[6pt]
&=n\left(2n^2+8n+10\right)\tag{5}
\end{align}
$$
A: 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 $4$.
