Proof that a Combination is an integer

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 $1$.

See my post here for a simple purely arithmetical proof that every binomial coefficient is an integer. The proof shows how to rewrite any binomial coefficient fraction as a product of fractions whose denominators are all coprime to any given prime $\rm\:p.\,$ This implies that no primes divide the denominator (when written in lowest terms), therefore the fraction is an integer.

The key property that lies at the heart of this proof is that, among all products of $\rm\, n\,$ consecutive integers, $\rm\ n!\ $ has the least possible power of $\rm\,p\,$ dividing it - for every prime $\rm\,p.\,$ Thus $\rm\ n!\ $ divides every product of $\rm\:n\:$ consecutive integers, since it has a smaller power of every prime divisor. Therefore $$\rm\displaystyle\quad\quad {m \choose n}\ =\ \frac{m!/(m-n)!}{n!}\ =\ \frac{m\:(m-1)\:\cdots\:(m-n+1)}{\!\!n\:(n-1)\ \cdots\:\phantom{m-n}1\phantom{+1}}\ \in\ \mathbb Z$$

As Jonas mentioned, it counts something so it has to be a natural number.

Another way is to notice that product of $m$ consecutive natural numbers is divisible by $m!$.(Prove this!)

So if we write $n! = n(n-1)(n-2) \cdots (k+1) \times (k!)$, we find that $k!$ divides $k!$ and

$n(n-1)(n-2) \cdots (k+1)$ is a product of $(n-k)$ consecutive natural numbers and hence $(n-k)!$ divides it.