Let $f\in Hol(\mathbb{C}\backslash\{a_1, \dots, a_N\})$ where ${a_1, \ldots ,a_N,\infty}$ are the poles of $f$. Show that $f$ is rational function.
I've tried to define $g(z):=\frac{f(z)}{\prod_{i=1}^{k=N}(z-a_k)}$ and show that $g$ has the form $\frac{1}{P(z)}$ where P(z) is polynomial
Let $p_1, \dots, p_N$ be the principal parts of $f$ at $a_1, \dots, a_N$, i.e. they are given by the negative coefficient-parts of the Laurent series expansion of $f$ at $a_i, i = 1, \dots, N$. Note that the $p_i$ are rational functions. Then $g = f - p_1 - \dots - p_N$ is entire, i.e. holomorphic on all of $\mathbb{C}$. Hence we can expand it as a Taylor series $$ g(z) = \sum_{k = 0}^{\infty} a_k z^k $$ Saying that $f$, and hence also $g$ has a pole (and not an essential singularity) at infinity means that $$ g\left(\frac1{z}\right) = \sum_{k=-\infty}^0 a_{-k} z^k $$ has a pole at $z = 0$, i.e. the series starts at some finite $-K$. But this means that the sum in the Taylor expansion for $g$ is finite, so $g$ is a rational function. Hence $f = g + p_1 + \dots p_N$ is a rational function.