Show that if $a$ and $b$ are positive integers then $(a!)^b b! \mid (ab)! $
This is what I have done:
The above statement is true for $a=1$ and any arbitrary value of $b$, and also for $b=1$ and any arbitrary value of $a$. Now for $a \ge 2 $ and $ b\ge 2$, $a+b\le ab $. This implies, $ (a+b)!\mid (ab)!$ and since $$\dbinom{a+b}{a} = \frac{(a+b)!}{a!b!} \implies a!b!\mid (a+b)! \implies (a!b!)\mid(ab)!$$
Any ideas how to further progress in this proof? Or are there any other ways to prove this?