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:


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

Note $S$ has $n$ elements.

By Set of Successive Numbers Contains Factor:


 * $\left\{ {m} \right\}$ contains a factor of $1$,


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

and in general,


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

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

Multiplying all elements of $S$ gives: