Show that $(a!)^b b! \mid (ab)! $

Question

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?

Answer 1

Induction on $b$ will work.

Base case: $b=1$ yields $(a!)^1\cdot1!\mid(a\cdot1)!$ which is true.

Inductive step assume for some positive integer $k$, we have $$(ak)!=n(a!)^k k!$$ where $n$ is a positive integer.

Consider $b=k+1$: \begin{align}(a(k+1))!&=(ak)!\cdot\frac{(ak+a)!}{(ak)!}\\&=n(a!)^k k!\binom{ak+a}{ak}\cdot a!\\&=n(a!)^{k+1}k!\cdot(k+1)\binom{ak+a-1}{ak}\\&=m(a!)^{k+1}(k+1)!\end{align} where $m$ is a positive integer.

Thus $(a!)^b b!\mid (ab)!$.