Ensuring the Multinomial Counting Theorem Produces Integer Coefficients

Question

As a result of the Multinomial Counting Theorem, we get that the number of ways to distinctly arrange $n$ objects with groups of identical objects each with length $a_1, a_2, ..., a_k$ is equal to:

$$\binom{n}{a_1,a_2,...,a_k} = \frac{n!}{a_1!a_2!\dots a_k!}$$

How do we ensure that this number is always an integer. In other words, given $n \in \mathbb{Z} \wedge (a_1,a_2,...,a_k) = \mathbb{Z}^k \wedge \sum a_i = n$, show that: $$\frac{n!}{a_1!a_2!\dots a_k!} \in \mathbb{Z}$$

The best "proof" I can think of is combinatorially, and it's rather intuitive because it's the number of ways to arrange certain objects, so it must be an integer. Is there an algebraic way to show this?

Answer 1

$\binom{n}{a_1,a_2,...,a_k} = \frac{n!}{a_1!a_2!\dots a_k!} = \binom{n}{a_1}\binom{n-a_1}{a_2}...\binom{(n-\sum_{i=1}^{k-1} a_i)}{a_k}$ and we know each is an integer.