Prove or disprove: For non-negative integers $m$ and $n$, $m!n! = (mn)!$

Question

I have rewritten the question as "If $m$ and $n$ are non-negative integers, then $m!n!$ = $(mn)!$"

Here is my current attempt. I am not sure if I am on the right path.

Proof.
Let $m$ and $n$ be non-negative integers.
So $m! = (m-1)! * m, n! = (n-1)! * n$, and $m!n! = mn(m-1)!(n-1)!$.
On the other hand, $(mn)! = (mn - 1)! * mn$.
...
Therefore, $m!n! = (mn)!$.

Am I headed in the right direction?

Answer 1

This statement can be disproven.

Counterexample.
Let $m = 0$ and $n = 5$.
So $m!n! = 0! * 5! = 1*5*4*3*2*1 = 120$.
On the other hand, $(mn)! = (0 * 5)! = 0! = 1$.
$120 ≠ 1$.
Therefore, $m!n! ≠ (mn)!$.

Answer 2

Assuming $m, n > 1$ then $m!n! < m!(m + 1).....(m + n) = (m + n)! \le (mn)!$.