I know that an analytic function on $\mathbb{C}$ with a nonessential singularity at $\infty$ is necessarily a polynomial.
Now consider a meromorphic function $f$ on the extended complex plane $\hat{\mathbb{C}}$. I know that $f$ has only finitely many poles, say $z_1,\dots,z_n$ in $\mathbb{C}$. Suppose also that $f$ has a nonessential singularity at $\infty$.
Then if $z_i$ have orders $n_i$, it follows that $\prod(z-z_i)^{n_i}f(z)$ is analytic on $\mathbb{C}$, and has a nonessential singularity at $\infty$, and is thus a polynomial, so $f$ is a rational function.
But I'm curious, what if $f$ doesn't have a singularity at $\infty$, or in fact has an essential singularity at $\infty$ instead? Is $f$ still a rational function?