Factorial Divides Product of Successive Numbers

Theorem
Let $m, n \in \N_{\ge 1}$ be natural numbers

Let $m^{\overline n}$ be $m$ to the power of $n$ rising.

Then:


 * $m^{\overline n} \equiv 0 \bmod n!$

That is, the factorial of $n$ divides the product of $n$ successive numbers.

Proof
Let $m \in \N_{\ge 1}$.

Consider the set:


 * $S = \set{m, m + 1, m + 2, \ldots, m + n - 1}$

Note $S$ has $n$ elements.

By Set of Successive Numbers contains Unique Multiple:


 * $\set m$ contains a factor of $1$


 * $\set {m, m + 1}$ contains a factor of $2$

and in general:


 * $\set {m, m + 1, \ldots, m + j - 1}$ contains a factor of $j$.

It follows that $S$ contains factors of $1, 2, 3, \ldots, n$.

Multiplying all elements of $S$ gives: