If a holomorphic function $f:\mathbb{C}\to\mathbb{C}$ is bounded, i.e. $|f| \lt A$ for some constant $A$ for the entire domain of $f$, then $f$ is constant according to Liouville's theorem.
Does this necessarily mean that any non-constant holomorphic function has all of $\mathbb{C}$ as its codomain?
Another way to phrase it: Does the equation $f(z) = v$ have a solution in $\mathbb{C}$ for any $v \in \mathbb{C}$?
Almost, but not quite. The example $f(z) = e^z$ shows that there are non-constant entire functions that omit a value. However, that is the most that can be the case, by Picard's (little) theorem, an entire function that omits (at least) two values is constant.
From a higher perspective, Picard's theorem is a consequence of the uniformisation theorem (together with Liouville's theorem), the universal (holomorphic) covering of the plane with two points removed is the unit disk, so an entire function that omits two values can be lifted to an entire holomorphic function with values in the unit disk, hence the lift is constant, and thus the function itself.
Of course Picard's theorem can be proved without the full power of the uniformisation theorem. (It's also a consequence of Picard's big theorem; let $z_0$ be an essential singularity of $f$, then $f$ assumes every complex value, with at most one exception, infinitely often in every punctured neighbourhood of $z_0$.)
