# 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 Unique Multiple:

- $\left\{ {m} \right\}$ 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:

\(\displaystyle m \left({m+1}\right)\ldots\left({m+n-1}\right)\) | \(=\) | \(\displaystyle k_1 2 k_2 3 k_3 \ldots \left({n-1}\right)k_{n-1} n k_n\) | $\quad$ for some $k_1, k_2, \ldots, k_n \in N$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({k_1 k_2 \ldots k_{n-1}k_n}\right)\left({1 \times 2 \times \ldots \times \left({n - 1}\right) \times n}\right)\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle m^{\overline n}\) | \(=\) | \(\displaystyle K n!\) | $\quad$ for some $K \in \N$ | $\quad$ | ||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle m^{\overline n}\) | \(\equiv\) | \(\displaystyle 0 \bmod n!\) | $\quad$ | $\quad$ |

$\blacksquare$