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:
 * $\displaystyle \forall m, n \in \Z_{>0}: \exists r \in \Z: \prod_{k \mathop = 1}^n \left({m+k}\right) = r \prod_{k \mathop = 1}^n k$

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: $\displaystyle \binom{m+n} m$.