$\begin{eqnarray*} D &=&% \begin{vmatrix} n! & (n+1)! & (n+2)! \\ (n+1)! & (n+2)! & (n+3)! \\ (n+2)! & (n+3)! & (n+4)!% \end{vmatrix} \\ &=&% \begin{vmatrix} n! & (n+1)n! & (n+2)(n+1)n! \\ (n+1)n! & (n+2)(n+1)n! & (n+3)(n+2)(n+1)n! \\ (n+2)(n+1)n! & (n+3)(n+2)(n+1)n! & (n+4)(n+3)(n+2)(n+1)n!% \end{vmatrix} \\ &=&n!^{3}(n+1)^{2}(n+2)\underset{\text{This determinant is 2 (see below)}}{\underbrace{% \begin{vmatrix} 1 & (n+1) & (n+2)(n+1) \\ 1 & (n+2) & (n+3)(n+2) \\ 1 & (n+3) & (n+4)(n+3)% \end{vmatrix}% }} \\ &=&n!^{3}\left( 2n^{3}+8n^{2}+10n+4\right) \end{eqnarray*}$
I have used the following property repeatedly: "If $B$ results from $A$ by multiplying one row or column with a number $c$, then $\det(B) = c \cdot \det(A)$". (Wikipedia). $n!$ is a factor of rows 1, 2 and 3. $n+1$ is a factor of rows 2 and 3. $n+2$ is a factor of row 3.
Thus
$\frac{D}{n!^{3}}-4=2n^{3}+8n^{2}+10n=(2n^{2}+8n+10)n.$
Evaluation of the last determinant expanding it by the minors of column 1.
$% \begin{vmatrix} 1 & (n+1) & (n+2)(n+1) \\ 1 & (n+2) & (n+3)(n+2) \\ 1 & (n+3) & (n+4)(n+3)% \end{vmatrix}% $
$=% \begin{vmatrix} (n+2) & (n+3)(n+2) \\ (n+3) & (n+4)(n+3)% \end{vmatrix}% -% \begin{vmatrix} (n+1) & (n+2)(n+1) \\ (n+3) & (n+4)(n+3)% \end{vmatrix}% +% \begin{vmatrix} (n+1) & (n+2)(n+1) \\ (n+2) & (n+3)(n+2)% \end{vmatrix}% $
$=(n+2)(n+3)% \begin{vmatrix} 1 & (n+3) \\ 1 & (n+4)% \end{vmatrix}% -(n+1)(n+3)% \begin{vmatrix} 1 & (n+2) \\ 1 & (n+4)% \end{vmatrix}% $
$+(n+1)(n+2)% \begin{vmatrix} 1 & (n+2) \\ 1 & (n+3)% \end{vmatrix}% $
$=(n+2)(n+3)-2(n+1)(n+3)+(n+1)(n+2)$
$=\left( n^{2}+5n+6\right) -\left( 2n^{2}+8n+6\right) +\left( n^{2}+3n+2\right) =2$