Prove that for every natural number $n\ge 2$, $n$ does not divide $n+1$.
Proof: Suppose for every natural number $n\ge 2$, $n$ does divide $n+1.$ However, for natural numbers $a$ and $b,$ $a$ divides $b$ or goes into $b$ if $b=ka$ for some natural number $k \ge 2$. Thus, there is only some $n$ that divides $n+1.$ Therefore, this is a contradiction.
A direct proof seems to be easier. If $n\geq 1$ divides $n+1$ then $n$ divides also the difference $n+1-n=1$ (note that $n$ divides $n$) which implies that $n=1$.
You may also say. If $n\geq 2$ divides $n+1$ then $n$ divides also the difference $n+1-n=1$ (note that $n$ divides $n$) which implies that $n=1$. Contradiction.