Show that $\displaystyle (n!+1, (n+1)!+1)=1$
if $\displaystyle d \mid n! + 1,$ then $\displaystyle d \mid (n+1)! + n+1.$ so if we also have $\displaystyle d \mid (n+1)! + 1,$ then $\displaystyle d \mid n$ and hence $\displaystyle d \mid n!,$ which completes the proof because we're given that $\displaystyle d \mid n!+1.$