VI OBM

The solutions are $n=1,k=1$; $n=2,k=1$; $n=4,k=2$. It is easy to check that the solutions above are the only solutions for $n\le 4$. So assume $n>4$. So $n!+1>n+1$, so $k>1$. If $n$ is odd, then $n+1$ is even, but $n!+1$ is odd, so there are no solutions. So $n$ is even. Hence $n$ is composite. So $n$ divides $(n-1)!$. But using the binomial theorem we have $(n+1{)}^k-1=n^k+(\genfrac{}{}{0ex}{}{k}{1})n^{k-1}+\cdots +(\genfrac{}{}{0ex}{}{k}{k-2})n^2+kn$. Hence $(n-1)!=n(n^{k-2}+n^{k-3}+\cdots +\left(\genfrac{}{}{0ex}{}{k}{k-2}\right))+k$. Hence $n$ divides $k$. But that means $k\ge n$, and $(n+1{)}^n>n!+1$. So there are no other solutions.