# A divisibility problem 2

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