# Factorial Divides Product of Successive Numbers

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## 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:

\(\ds m \paren {m + 1} \dotsm \paren {m + n - 1}\) | \(=\) | \(\ds k_1 2 k_2 3 k_3 \dotsm \paren {n - 1} k_{n - 1} n k_n\) | for some $k_1, k_2, \ldots, k_n \in N$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {k_1 k_2 \dotsm k_{n - 1} k_n} \paren {1 \times 2 \times \dotsm \times \paren {n - 1} \times n}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds m^{\overline n}\) | \(=\) | \(\ds K n!\) | for some $K \in \N$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds m^{\overline n}\) | \(\equiv\) | \(\ds 0 \bmod n!\) |

$\blacksquare$