Problem: Show that if $f$ is entire and nowhere zero, then there exists an entire function $g$ such that $f(z) = g(z)^2$ for all $z\in \mathbb{C}$.
Preliminary: To get some practice in, I tried to solve an analogous problem showing that there exists an entire $g$ such that $f=e^g$:
To find such $g$: we different both sides to get $f' = g' e^g = g' f \Rightarrow \frac{f'}{f} = g'$. Since entire functions on a simply connected space have an antiderivative, this implies that $f = e^{g + c} = Ce^g$ for some constant $C$. So there is an entire function $g$ such that $f = e^g$ as desired.
Attempt on the original problem: I seem unable to apply the same technique as the $f=e^g$ proof. That is, if $f=g^2$ then $f' = 2gg'$ so we have to stop here. Furthermore, considering $\sqrt{f(z)} = g(z)$ seem illegal because we have to consider a branch cut: this is a problem because Picard's little theorem tells us that if $f$ is non-constant entire that doesn't hit the value $0$, then the range is $f(\mathbb{C}) = \mathbb{C} \setminus \{ 0 \}$ so there isn't an appropriate branch cut we can really take.
What can I do here? One intuitive guess is if I know $f = e^h$ for some entire $h$, then I can consider $g = e^{h/2}$ so that $g^2 = e^h = f$ and this helps us avoid the "square-root branch cut" issue?
Following @aschepler 's comments, here is the solution:
Suppose $f$ is entire and nonzero. Then, $f'$ is also an entire function. Since $f$ is nonzero, we can take the quotient to get another entire function $f'/f$. Since this entire function is defined on $\mathbb{C}$ (by definition) which is simply connected, it has an antiderivative. Consequently there exists an entire $h$ such that $f = e^h$.
Therefore, $g = e^{h/2}$ is an entire function since $h/2$ is entire and the composition of two entire functions (namely $h/2$ and $e^z$) is entire. Observe that $g^2 = e^h = f$ so we are done.