Use properties of determinant and show 
Let $n$ be a positive integer and 
\begin{align} M =
    \begin{pmatrix}
    n! & (n+1)! & (n+2)! \\
    (n+1)! &(n+2)! & (n+3)! \\
    (n+2)! & (n+3)! & (n+4)! \\
    \end{pmatrix}
\end{align}
Use properties of determinant to show that \begin{align}\left(\frac{|M|}{(n!)^3}- 4\right)\end{align} is divisible by $n$.

I took an $n!$ factor out of the matrix, getting a new matrix $B$ such that $\det A = n^3 \det B$, since $A$ and $B$ are row equivalents. But what is $\det B$ ?
 A: So, they're looking for the determinant of $M$.  We can make that easier to calculate by factoring a common term of $n!$ out of each row, giving us a new matrix:
$$N=\left[\begin{array}{ccc}
1 & n+1 & (n+1)(n+2)\\
n+1 & (n+1)(n+2) & \small{(n+1)(n+2)(n+3)}\\
(n+1)(n+2) & \small{(n+1)(n+2)(n+3)} & \scriptsize{(n+1)(n+2)(n+3)(n+4)}
\end{array}\right]$$
Now, $|N|=\frac{|M|}{(n!)^3}$, so we just need to find $|N|-4$.  It looks like we're about to dive into horrible polynomial multiplications.  But wait!  We only need to show that $|N|-4$ is divisible by $n$.  That means that if the constant term of the polynomial $|N|$ is 4, we're done.  We can do that by multiplying those terms out and ignoring all but the constant term"
$$\left[\begin{array}{ccc}
1 & ...+1 & ...+2\\
..+1 & ...+2 & ...+6\\
...+2 & ...+6 & ...+24
\end{array}\right]$$
Now the determinant is easy enough to calculate.  Using Sarrus' Rule, the value of this determinant is 
$$(1\cdot2\cdot24) + (1\cdot6\cdot2) +(2\cdot1\cdot6)- (2\cdot2\cdot2) - (1\cdot6\cdot6) - (1\cdot1\cdot24) $$ $$ = 48+12+12-8-36-24=4$$.
Therefore, $|N|=An^6+Bn^5+Cn^4+Dn^3+En^2+Fn+4$ for some integer values of $A,B,C,D,E,F$.  We don't need to find those values, though.  The only thing that matters is that every term of $|N|-4$ contains a power of $n$, which is what we were asked to show.
A: After you take the common factor $n!$ out of all columns, take $(n+1)$ and $(n+1)(n+2)$ out of 2nd and 3rd columns, respectively:
$$\frac{|M|}{(n!)^3}=\left|\begin{array}{ccc}
1 & \color{red}{n+1} & \color{blue}{(n+1)(n+2)}\\
n+1 & (\color{red}{n+1})(n+2) & \small{\color{blue}{(n+1)(n+2)}(n+3)}\\
(n+1)(n+2) & \small{(\color{red}{n+1})(n+2)(n+3)} & \scriptsize{\color{blue}{(n+1)(n+2)}(n+3)(n+4)}
\end{array}\right|=\\
\color{red}{(n+1)}\color{blue}{(n+1)(n+2)}\left|\begin{array}{ccc}
1 & 1 & 1\\
n+1 & n+2 & \small{n+3}\\
(n+1)(n+2) & \small{(n+2)(n+3)} & \scriptsize{(n+3)(n+4)}
\end{array}\right|\stackrel{C_3-C_2\to C_3\\ C_2-C_1\to C_2}{=}\\
(n+1)^2(n+2)\left|\begin{array}{ccc}
1 & 0 & 0\\
n+1 & 1 & 1\\
(n+1)(n+2) & 2n & 2n+2
\end{array}\right|\stackrel{C_3-C_2\to C_3}=\\
(n+1)^2(n+2)\left|\begin{array}{ccc}
1 & 0 & 0\\
n+1 & 1 & 0\\
(n+1)(n+2) & 2n & \color{red}2
\end{array}\right|=\color{red}2(n+1)^2(n+2)$$
Hence:
$$\begin{align}\left(\frac{|M|}{(n!)^3}- 4\right)\end{align}=2n^3+8n^2+10n\equiv 0 \pmod{n}.$$
