If $n\in \mathbb{N}$ and $n\ge 2$, then the numbers $\left[n!+2,n!+3,n!+4,\dots ,n!+n\right]$ are all composite. Thus for any $n\ge 2$, 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 $1.

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 $1. All that remains is to show that $b\in \mathbb{N}$ and $1:

• $b=\left(\frac{n!}{a}+1\right)\in \mathbb{N}$ because $a\le n\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}a|n!$
• $ab=c$ and $1, therefore $1

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

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 $\left[n!+2,n!+3,n!+4,\dots ,n!+n\right]$ 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. $◻$

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