Solution
= Solution
Suppose $g(\mathbb C)$ is not dense. Then some <open disc> $D(w,r)$ is disjoint from the image, so
$$
F(z)=\frac1{g(z)-w}
$$
is entire and satisfies $|F(z)|\leq1/r$. Liouville's theorem makes $F$, and hence $g$, constant, a contradiction. Therefore every nonconstant entire $g$ has dense image, as asserted by <dense image of a nonconstant entire function>.