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.

Denote $s_j = m + j$ for $j = 0, 1, \ldots, n-1$.

By the Division Theorem, for each $s_j \in S$, there are $k_j, r_j \in \Z$ such that:


 * $s_j = j k_j + r_j$, where $0 \le r_j < j$

By the Pigeonhole Principle, $r_j = 0$ for some element of $S$ for each $j = 0, 1, \ldots, n - 1$.

Multiplying all elements of $S$ gives: