 Solution

Suppose $g(\mathbb{C})$ is not dense. Then some open disc $D(w,r)$ is disjoint from the image, sois entire and satisfies $|F(z)|\le 1/r$. 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.