Use the Mittag-Leffler theorem to prove the following: Let $(a_n)$ be a sequence in a simply connected domain $D \subset \mathbb{C}$ that does not have an accumulation point in $D$. Prove that there exists a holomorphic function $f$ so that its zero set equals $(a_n)$ counting multiplicity.
In particular, do not use the Weierstrass Factorization Theorem, Weierstrass Products, or Blaschke Products in your answer.
I have no idea how to do this problem. Can someone help?
Idea: If $f(a) = 0$ with multiplicity $k$ then the logarithmic derivative $(\log f)' = f'/f$ satisfies $$ \frac{f'(z)}{f(z)} = \frac{k}{z-a} + O(1) \quad \text{ for } z \to a \, . $$
Use the Mittag-Leffler theorem to construct a meromorphic function $g$ in $D$ with poles exactly at the points $a_n$ and principal parts $$ \frac{k_n}{z-a_n} $$ where $k_n$ is the desired multiplicity at $a_n$.
Then show that there is a holomorphic function $f$ in $D$ such that $$ \frac{f'}{f} = g \, ,$$ details can be found for example in Which meromorphic functions are logarithmic derivatives of other meromorphic functions?.
$f$ has zeros exactly at the points $a_n$ with multiplicity $k_n$.