49th Mathematical Olympiad in Ukraine

If $k\ge 2$ then the sum $(k!+2)$ is divisible by $2$, and so $n=2m$. Then for $k\ge 4$ in the equality $(2m{)}^3-k!=2$ the left part is divisible by $4$, but the right one is not. Thus, all that we need is to consider the cases $k=1,2,3$. An easy check shows that there is an only solution $n=2,k=3$.