Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function. How can one show that if $f$ is not a polynomial, then $\overline{f(A)}=\mathbb{C}$, where $A$ is the annulus $\{z\in\mathbb{C}: |z|>r\},\forall r>0$?
If $f$ is a polynomial, then the funtion $g:\mathbb{C}\backslash\{0\}\rightarrow \mathbb{C}$, given by $z\mapsto g(z)=f(1/z)$, has a pole at $z=0$ (and vice-versa). Does this help?
Hints: Suppose $f(A)$ is not dense. Then there exists an open disk $B(w,s)$ disjoint from $f(A)$. Note that $|\frac 1{f(z)-w}| \leq \frac 1 s$ for $|z| >r$. Let $z_1,z_2,...,z_n$ be the zeros of $f(z)-w$ in $\{z:|z| \leq r\}$ counted according to multiplicities. Let $g(z)=\frac {\prod_j (z-z_j)} {f(z)-w}$. Then $g$ is an entire function and there exists a finite constants $C,D$ such that $|g(z)| \leq C|z|^{n}+D$. This implies that $g$ is a polynomial of degree at most $n$. Can you finish?