# For all $n>2$ there exists a prime number between $n$ and $n!$

How to prove that there exists a prime number between $n$ and $n!$, for all $n> 2$?

(Bertrand's postulate gives a much better bound, but this question is about obtaining a self-contained proof.)

• Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas?
– 5xum
Dec 29 '14 at 7:31
• For $n\geq 3$, take $p$ as a prime divisor of $n!-1$. Dec 29 '14 at 8:02

Given $n>2$. So for every integer $x$ such that $1<x<(n+1),$ we have $x|n!$ and $x\not|(n!-1).$

$\therefore$ either $(n!-1)$ is a prime, or $\exists$ a prime $p\ge (n+1)$ such that $p|(n!-1)$.

So in any case, $\exists$ a prime $p$ such that $(n+1)\le p\le (n!-1)$.

• And what if $n!-1$ is a prime. This is possible for $n=3, 4$. Just a few small examples. Could you please explain?
– Paul
Feb 6 at 16:24
• @Paul For any $n>2$ if $n!-1$ is a prime, then you can stop the search for existence of a prime between $n$ and $n!$. Try writing the usual ordering of the numbers being discussed here. Feb 6 at 16:51

For $n=2$ take $p=2$. For $n\geq 3$, take $p$ as a prime divisor of $n!-1$ (hence $p<n!$). Then $p>n$, if not, $p$ divide $n!$, hence divide $n!-(n!-1)=1$.

• Why does p divide $n! - (n!-1)$? Sep 17 '16 at 8:49
• @Donald because if $p$ divides $n!-1$, and $p$ divides $n!$, then $p$ must divide their difference, namely, $p$ divides $n!-(n!-1)$. Aug 3 '20 at 23:56

Base case: $3<5<3!$

Induction Hypothesis: Suppose that $\exists p$ such that $p$ is prime and $(n-1)<p<(n-1)!$

Induction Step: We must now show that $\exists k$ such that $k$ is prime and $n<k<n!$.

So either $n=p$ or $n<p<(n-1)!<n!$ If the case is the latter one, we are done. So, suppose that $n=p$. Now, we consider $Q=\prod p_i$ where $p_i\leq p$ and $p_i$ is prime. Well, we actually consider $Q+1$. Notice that $p<Q+1<n!$ Furthermore, notice that no primes less than or equal to $p$ divide $Q+1$ (because of the left over 1). Hence, either $Q+1$ is prime or composite. If $Q+1$ is prime, we are done. If $Q+1$ is composite (and since no prime less than or equal to $p$ divides $Q+1$) there must exist some $q<t<Q+1<n!$ such that $t|(Q+1)$. Thus, we are finished by setting $k=Q+1$ or $k=t$.