Looking for a non-combinatorial proof that $a! \cdot b! \mid (a+b)!$ (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?
 A: If we use as a lemma that the product of $k$ consecutive integers is divisible by $k!$ (proven e.g. here: The product of n consecutive integers is divisible by n factorial), we see that $(a+b)!/a!$ factors into the $b$ consecutive integers $a+1,\ldots,a+b$, hence is divisible by $b!$. Then $a!b!|(a+b)!$.
A: Possibly the answer that you don't want:
$$\frac{(a+0)!}{a!\cdot 0!}=1=\frac{(0+b)!}{0!\cdot b!}$$
and
$$\frac{(a+b)!}{a!\cdot b!}=\frac{(a+b)(a+b-1)!}{a!\cdot b!}=\frac{(a-1+b)!}{(a-1)!\cdot b!}+\frac{(a+b-1)!}{a!\cdot (b-1)!}.$$
Then by induction, Pascal's triangle is made of integers.
A: With the settings
\begin{align*}
[k]_q:=\frac{1-q^k}{1-q}\qquad\text{and}\qquad [k]_q!:=\prod_{j=1}^{k}[j]_q=\prod_{j=1}^{k}\frac{1-q^j}{1-q}
\end{align*}
the q-binomial coefficient $\begin{bmatrix}a+b\\a\end{bmatrix}_q$ is defined as
\begin{align*}
\begin{bmatrix}a+b\\a\end{bmatrix}_q:=\frac{[a+b]_q!}{[a]_q![b]_q!}
\end{align*}

Since the following is valid
  \begin{align*}
\begin{bmatrix}a+b\\a\end{bmatrix}_q&=\begin{bmatrix}a+b-1\\a\end{bmatrix}_q+q^{b}\begin{bmatrix}a+b-1\\a-1\end{bmatrix}_q\\
\begin{bmatrix}a\\0\end{bmatrix}_q&=\begin{bmatrix}a\\a\end{bmatrix}_q=1
\end{align*}
  the $q$-binomial coefficients are polynomials in $q$ with non-negative integer coefficients.
Together with
  \begin{align*}
\frac{(a+b)!}{a!b!}=\lim_{q\rightarrow 1}\begin{bmatrix}a+b\\a\end{bmatrix}_q
\end{align*}
the claim $a!b!|(a+b)!$ follows.

A: There's a truly beautiful proof by Tim Gowers.
And a similar question has been asked before.
A: You can use $\nu_{p}(n!)=\sum \limits_{k\ge 1}\left[\dfrac{n}{p^k}\right]$ and that $[a+b]\ge [a]+[b]$
