Divisibility of Product of Consecutive Integers

Theorem
The product of $$n$$ consecutive positive integers is divisible by the product of the first $$n$$ consecutive positive integers.

That is:
 * $$\forall m \in \Z_+^*: \exists r \in \Z: \prod_{k=1}^n \left({m+k}\right) = \prod_{k=1}^n k$$

For example:
 * $$5 \cdot 6 \cdot 7 \cdot 8 = 1680 = 70 \cdot 24 = 70 \cdot 1 \cdot 2 \cdot 3 \cdot 4$$


 * $$10 \cdot 11 \cdot 12 \cdot 13 = 17160 = 715 \cdot 24 = 715 \cdot 1 \cdot 2 \cdot 3 \cdot 4$$


 * $$4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 6720 = 56 \cdot 120 = 56 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5$$


 * $$11 \cdot 12 \cdot 13 \cdot 14 \cdot 15 = 360360 = 3003 \cdot 120 = 3003 \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5$$

Proof
$$ $$ $$ $$

Hence the result, and note that for a bonus we have identified exactly what the divisor is: $$\binom{m+n} m$$.