Proof that a Combination is an integer From its definition a combination $\binom{n}{k}$, is the number of distinct subsets of size $k$ from a set of $n$ elements.

This is clearly an integer, however I was curious as to why the expression

$$\frac{n!}{k!(n-k)!}$$ always evaluates to an integer.

So far I figured:

$n!$, is clearly divisible by $k!$, and $(n-k)!$, individually, but I could not seem to make the jump to proof that that $n!$ is divisible by their product.
Well, one noncombinatorial way is to induct on $n$ using Pascal's triangle; that is, using the fact that ${n \choose k} = {n-1 \choose k - 1} + {n-1 \choose k}$ (easy to verify directly) and that each ${n - 1 \choose 0}$ is just $1$.
One way to get there: any product of $k$ consecutive positive integers $d := \frac{n!}{(n-k)!} = \prod^k (n-i+1)$ is divisible by $k!$.

You can assure yourself of this by considering divisibility by the primes and prime powers up to $k$. For example, $k!$ has $\lfloor k/p\rfloor$ multiples of $p$, and $d$ will have either the same number of multiples of $p$ or one more, since the "first" multiple of $p$ in $k!$ will be at $p$, whereas it may be earlier in $d$.

The quantity I have denoted by $d$ is the $k$-descending factorial of $n$, sometimes written $n^{\underline k}$ or $(n)_k$.