Suppose $f$ is entire and $\lim_{z\to\infty}f(z)=\infty$. Show that $f(\mathbb{C})=\mathbb{C}$

The niceties of the one-point compactification of $\Bbb C$ aside, consider:

$f(z) \ne w, \; \forall z \in \Bbb C; \tag 1$

$f(z) - w \ne 0, \; \forall z \in \Bbb C; \tag 2$

$(f(z) - w)^{-1}$ is then entire; since

$f(z) \to \infty \; \text{as} \; z \to \infty \tag 3$

$(f(z) - w)^{-1}$ is bounded; hence by Liouville's theorem, constant; but then we cannot have (3); thus (1) fails.

The statement $\lim_{z\to\infty}f(z)=\infty$ means $|f(z)|\to \infty$ as $|z|\to\infty$. It can be seen as convergence to $\infty \in\Bbb C_\infty$ where $\Bbb C_\infty$ is a one-point compactification of $\Bbb C$. In fact, this property implies that $f(z)=p(z)$ for some non-constant polynomial $p$. By fundamental theorem of algebra, it follows $f(\Bbb C)=\Bbb C$. To see this, let $ g(z)=f(\frac1{z}) $ for $z\ne 0$. Since $\lim_{z\to 0}|g(z)|=\infty$ (which is true because as $z\to 0$, $1/z \to \infty$ and $g(z)=f(1/z)\to\infty$ by the assumption), it follows that $g$ has an $n$-th pole at $0$, i.e. for some $n\ge 1$, $a_{-n}\ne 0$ and for all $z\ne 0$, $$ g(z)=\sum_{k=-n}^\infty a_kz^k. $$ (Here, $g$ having a pole is a classical result. Since $1/g$ tends to $0$ as $z\to 0$, it has a removable singularity at $0$, which is also an $n$-th zero. It follows $1/g(z)=z^nh(z)$ for some $h\ne 0$ on a neighborhood of $0$. So $g(z) = z^{-n}\frac1{h(z)}$ has an $n$-th pole.) Thus $f(z)=g(\frac1{z})=\sum_{k=0}^n a_{-k}z^k +\sum_{i>0}a_{i}z^{-i}$, and since $f$ is entire, it follows $$f(z)=\sum_{k=0}^n a_{-k}z^k=p(z).$$

Let $g(z):=f(1/z)$ for $z \ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) \to \infty$ as $z \to 0$, $g$ has a pole at $0$.

Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is given by

$g(z)=\sum_{n=0}^{\infty}a_n \frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.

Hence $f$ is a non- constant poynomial. Now use the fundamental theorem of algebra.