Prove that for $a,b \in \mathbb{N}$, $a!b!$ divides $(a+b)!$. Prove that for $a,b \in \mathbb{N}$, $a!b!$ divides $(a+b)!$.


This follows since $\binom{a+b}{a} = \dfrac{(a+b)!}{a!b!}$, which is an integer.

Are there any other arguments besides this one to prove the result or do I need to elaborate on the one above?
Your argument is fine and is the simplest one.

For another argument, this follows from Lagrange's theorem in group theory because $S_a \times S_b$ can be seen as a subgroup of $S_{a+b}$.
$C(a,b)$ is $(a+b)!/(a!b!)$ Which means the number of ways to select $1,2,\dots,a$ from $1,2, \dots,a+b$ without permutation therefore it is always a integer