Verify that (n!+1,(n+1)!+1)=1, problem worth trying

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 $d$

$h(n+1)-k=\frac{n}{d}$

$\Rightarrow\frac{n!}{d}+\frac{1}{d}=1$

$\Rightarrow{d=1}$.