(I use $a$ and $b$ to denote natural numbers.)
Question. Without appealing to the combinatorial interpretation of $$\frac{(a+b)!}{a! b!}$$ as a multinomial coefficients, is there a proof that for all $a$ and $b$, we have $$a! \cdot b! \mid (a+b)! \qquad?$$
Basically, I want a proof that just uses some clever algebra.
I was thinking that maybe we can use modular arithmetic, and try to understand the value of $(a+b)!$ modulo $a! \cdot b!$, and eventually show that this is $0$.
Ideas, anyone?