Let $E = F (u),$ where $u$ is algebraic over $F$, of degree $m.$ If $k$ is a natural with $(k, m) = 1,$ it is true that $E = F(u^{k})$? I tried but I couldn't.

I did another exercise that is reasonably similar. Let $E = F (u),$ where $u$ is algebraic over $F$, of odd degree. So $E = F(u^{2}).$ I tried to imitate the demonstration of this exercise, but I couldn't.