holomorph vs having a primitive

Question

Let $U \subseteq \mathbb{C}$ be an open set, and $f : U \to \mathbb{C}$

I know that if $f$ is holomorph, then $f'$ is holomorph, and so derivatives of all orders exists.

I know that beeing holomorph is not enough to have a primitive, for example $f(z) = 1/z$ is holomorph but does not have a primitive in $\mathbb{C} - \{0\}$

My question is, if $f$ has a primitive, let's say $F$ (so that $F' = f$), then does it mean that $F$ has a primitive too? (so if it have a primitive, it has primitives of all orders). If it's true, how is it proven? If it isn't, which function is a counterexample?

Thanks!

Answer 1

You have it at your disposal! For any function $f$ which doesn't have a primitive, take its derivative. Voila!