If $n \in \mathbb{N}$ and $n \geq 2$, then the numbers $[n! + 2, n! + 3, n! + 4, \ldots , n! + n]$ are all composite. Thus for any $n \geq 2$, we can find $n - 1$ consecutive composite numbers. The implication here being that there are arbitrarily large “gaps” between prime numbers.

I came up with an arithmetic proof which utilizes the fact that any composite number $c$ can be written as $c = ab$ where $a,b \in \mathbb{N}$ and $% $.

Proof. Assume $C$ is the set of all consecutive numbers following $n! + 1$ such that:

For any $c \in C$ we can write:

We already know that $a \in \mathbb{N}$ and $% $. All that remains is to show that $b \in \mathbb{N}$ and $% $:

• $b = (\frac{n!}{a} + 1) \in \mathbb{N}$ because $a \leq n \implies a \vert n!$
• $ab = c$ and $% $, therefore $% $

Therefore any $c \in C$ is composite. $\Box$

I realized that I took the bumpy route when I read the wikipedia article on Prime Gaps, which has a much simpler proof.

Proof. In the sequence $[n! + 2, n! + 3, n! + 4, \ldots , n! + n]$ the first term is divisible by $2$, the second term by $3$ and so on. Thus this is a sequence of $n - 1$ consecutive composite integers. $\Box$

This one-liner works because all numbers from $1 \ldots n$ are divisors of $n!$ 🤦🏽‍♂️