This is the first exercise from Sierpinski's Elementary Theory of Numbers. He gives a proof using induction and I was wondering if this approach was correct as well:
$a!b!|(a+b)! \iff \exists c \in \mathbb{N} \text{ such that } (a+b)! = c(a!b!)$
Assuming without loss of generality that $a \leq b$:
$a!b! = \displaystyle\prod\limits_{n=1}^a n^2 \displaystyle\prod\limits_{n=a+1}^b n$
Then we define the set S:
$S = \{n \in \mathbb{N} :n < a^2 \wedge \not \exists m \in \mathbb{N}\text{ such that }m^2=n) \} $
If $c = \displaystyle\prod\limits_{n \in D}n \displaystyle\prod\limits_{n=b+1}^{a+b}n$,
then $(a+b)! = c(a!b!)$