Let $n=p$ - prime. Then $k^n+1=k+1\mod p$. Therefore $p\not |k+1$ for any odd prime or $k=2^l-1$. Let $k=2^l+1$ and n is odd. For odd prime p we denote $q(p)$ minimal odd m, suth that $p|k^m+1$. If not exist suth m, then $q(p)=\infty$. $n|k^n+1\to \forall p|n \ q(p)|n$. Let $p|n$ suth prime divisor, that $q(p)$ is minimal. If p is minimal, then $q(p)=1$ or $p|k+1=2^l$ - contradiction. If p is not minimal, then for $p_0<p, p_0|n$ we get $q(p_0)=\infty$. Therefore $k=2^l-1$ work for all $l\ge 2$.