2019/008) Show that $n^2(n^2-1)(n^2-4)$ is divisible by 360

Proof:
We have $n^2(n^2-1)(n^2-4)= n^2(n-1)(n+1)(n-2)(n+2) = (n-2)(n-1)n^2(n+1)(n+2)$
$=5! * {n\choose 5}$

Now $5!=120$ Additionally one of (n-2),(n-1),n is divisible by 3.
If n is divisible by 3 then $n^2$ is divisible by 9 or 2 number (n-2) and (n+1) or (n-1) and (n+2) are divisible by 3.
So the product by $3^2$ so the number is divisible by $LCM(120,9)$ or 360
Or alternatively there is an additional 3 so the product is divisible by 120 *3 or 320