Show that if $f(z)=w$ has at most $n$ solutions for each $w \in \mathbb{C}$ then the pole $z_0$ has order at most $n$ 
Let $D \subset \mathbb{C}$ be open and connected, $z_0 \in D$ and $n \in \mathbb{N}$. Suppose that $f : D \ \backslash \{z_0\} \to \mathbb{C}$ is holomorphic in $D \ \backslash \{z_0\}$ with the following property:
$$f(z) = w \ \text{has at most} \ n \ \text{solutions for all} \ w \in \mathbb{C}.$$
Show that $z_0$ is either a removable singularity or a pole of $f$
of order less than or equal to $n$.

This is a question that I could not complete in a recent complex analysis example sheet. So I need to prove that $z_0$ cannot be an essential singularity or a pole of order greater than $n$. I managed to prove that it cannot be essential using the Casorati-Weierstrass Theorem. But despite a lot of effort I couldn't show that $z_0$ cannot be a pole of order greater than $n$. I'm not sure if this is the best way to approach it, but taking the contrapositive we have the equivalent statement:
If $g: D \to \mathbb{C}$ is holomorphic, $g(z_0) \neq 0$, $m>n$, then there exists $w \in \mathbb{C}$ such that the equation $g(z)=w(z-z_0)^m$ has more than $n$ solutions.
I hope this is correct. But regardless I couldn't find any tools that would help to prove such a statement. Help would be appreciated.
 A: Assume wlog $g$ has a pole of order $m \geq 2$ at $0$, with coefficient $c \neq 0$; let $q(z) = z^{m-1}(g(z)-\frac{c}{z^m})$ analytic on a disc of radius $2r$ near $0$, fixing such an $r$ and assume $|q(z)| \leq A, |z| \leq r$ and pick any large enough $B$ s.t. $Br^m > Ar + |c|$; then it is obvious that $Bz^{m}, Bz^{m}-zq(z)-c$ have the same number of roots, namely $m$ on the disc with radius $r$ by Rouche and our choice of $B$, while since $c \neq 0$ none of the roots of $Bz^{m}-zq(z)-c$ can be zero and actually they are all distinct for almost all $B$ - actually by choosing $r$ small enough s.t. $m|c| > |z^2q'(z)|+(m-1)|zq(z)|, |z| \leq r$, which we always can do since the RHS of the inequality goes to zero with $|z|$, we can insure no double roots for all $B$ large enough to satisfy the required inequality as a simple computation shows.
But now $Bz^{m}-zq(z)-c = 0, z \neq 0$ means $g(z)=B$ since by definition $g(z) = \frac{c}{z^m}+\frac{q(z)}{z^{m-1}}$, so we are done showing that $g$ having a pole of order $m \geq 2$ means $g$ takes all large enough values $m$ times.
Note that for all analytic or meromorphic functions, double roots of any $g-B$ are isolated since they are zeros of the $g'$, so as noted above the extra argument above with small $r$ etc is not really needed but I included it for completeness
