Proposition. Let $f:\hat{\bC}\to \hat{\bC}$ be a meromorphic function. Then $f$ is a rational function.
To prove this, we'll use some lemmas.
Lemma 1. Let $f$ be such a meromorphic function. Then it has only a finite number of poles.
Proof. A meromorphic function is just a holomorphic function in $\hat{\bC}$ (minus a discrete set of poles). If $a=\infty$ then, since the set $S$ of singularities is discrete and closed, then it's discrete and compact, thus finite.
Proof of Proposition. Let $a_1,\cdots, a_n$ be the poles of $f$ and take
$$g(z)=\prod_{i=1}^n (z-a_i)^{o(a_i)} f(z).$$
Then $f(z)$ is holomorphic in $\bC$, and still meromorphic in $\hat{\bC}$. So,
$$\lim_{z\to\infty} g(z)=\in \hat{\bC}.$$
If $\lim_{z\to\infty} g(z)\in \bC$, then by Liouville theorem $g$ is constant and $f(z)=\frac{g(12)}{\prod_{i=1}^n(z-a_i)^{o(a_i)}}$.
If $\lim_{z\to\infty} g(z)=\infty$ then $\infty$ is a pole of $g$, and by the lemma shown here, $g$ is a polynomial. In both cases we're done.
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