Prove that if $p\le n$, then $p$ does not divide $n! + 1$

Question

I'm having trouble on how to approach this problem

Prove that if $p\le n$, then $p$ does not divide $n! + 1$ ($p$ is prime and $n$ is an integer).

its trivial, $p$ divides $n!$, so it can't divide $n!+1$ at the same time — Mirko, Nov 24 '14 at 02:19
$d\le n\Rightarrow d\mid n!.,$ so $\bmod d!:\ n!\equiv 0,$ so $,\color{#c00}{n!+1\equiv 1},,$ so $,d\mid \color{#c00}{n!+1}\iff d\mid\color{#c00} 1,,$ by divisibility mod reduction $\ \ $ — Bill Dubuque, Jun 30 '25 at 15:37

score 5 · Accepted Answer · answered Nov 24 '14 at 02:19

5

$p$ divides $n!$. So it cannot divide $n!+1$.

answered Nov 24 '14 at 02:19

rewritten

1 Answers1