# Factorial Divides Product of Successive Numbers

 It has been suggested that this page or section be merged into Divisibility of Product of Consecutive Integers. (Discuss)

## 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.

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

 $\ds m \left({m+1}\right)\ldots\left({m+n-1}\right)$ $=$ $\ds k_1 2 k_2 3 k_3 \ldots \left({n-1}\right)k_{n-1} n k_n$ for some $k_1, k_2, \ldots, k_n \in N$ $\ds$ $=$ $\ds \left({k_1 k_2 \ldots k_{n-1}k_n}\right)\left({1 \times 2 \times \ldots \times \left({n - 1}\right) \times n}\right)$ $\ds \implies \ \$ $\ds m^{\overline n}$ $=$ $\ds K n!$ for some $K \in \N$ $\ds \implies \ \$ $\ds m^{\overline n}$ $\equiv$ $\ds 0 \bmod n!$

$\blacksquare$