Simple 2

From combinatorics, $\binom{n}{r} \in \mathbb{Z^{+}}$ for every $n,r \in \mathbb{Z}$ and $0 \leq r \leq n$ [Since the enumeration of picking $r$ things out of $n$ things must be a positive integer].

Now, $\binom{n}{r} \in \mathbb{Z^{+}} \Rightarrow \dfrac{\prod_{j=0}^{r-1}(n-j)}{r!} \in \mathbb{Z^{+}}$ , i.e. product of $r$ consecutive natural numbers is divisible by $r!$ .

we can proof this by mathematical induction