TITLE: Looking for a non-combinatorial proof that $a! \cdot b! \mid (a+b)!$
QUESTION [6 upvotes]: (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?

REPLY [2 votes]: There's a truly beautiful proof by Tim Gowers.
And a similar question has been asked before.