Verify that $(n!+1,(n+1)!+1)=1$
Let $(n!+1,(n+1)!+1)=d$
Then there exists $(h,k)=1$ such that
$n!+1=dh$.....................(i)
$n!=dh-1$ and
$(n+1)!+1=dk$...............(ii)
From (ii) let
$(n+1)n!+1=dk$
$\Rightarrow{(n+1)(dh-1)}+1=dk$
By expansion of brackets, we get 
$ndh-n+dh-1+1=dk$
$ndh-n+dh-dk=0$
$ndh+dh-dk=n$
$d(nh+h-k)=n$
$d[(n+1)h-k)=n$
Divide through by $d$
$h(n+1)-k=\frac{n}{d}$
$\Rightarrow\frac{n!}{d}+\frac{1}{d}=1$
$\Rightarrow{d=1}$.