$f$ has a zero of order $m\iff \frac{1}{f}$ has a pole of order m Question

Let $f$ be holomorphic in a domain $D\subset \Bbb{C}$. Then $f$ has a zero of order $m$ in $z_0\in D \iff \frac{1}{f}\in H({D \setminus f^{-1}(0)}) \text{ has a pole of order $m$ in } z_0$.

My attempt:
I have proved the "$\implies$" direction.
For the other implication, we suppose that $$\min\left\{v\in \Bbb{N} : \frac{(z-z_0)^v}{f}\text{ is bounded near }z_0\right\}=m$$
We need to find a $g\in H(D)$ with $g(z_0)\neq 0$ such that $f = (z-z_0)^m g$. 
I haven't been able to do this. Please tell me what I could do.

Hint: If $\frac{1}{f}$ has a pole $z_0$ of order $m$ then there exists a function $\varphi (z)$ analytic and nonzero at $z_0$ such that $$\frac{1}{f(z)} = \frac{\varphi (z)}{(z- z_0)^m}$$ then $$f(z) = (z-z_0)^m \frac{1}{\varphi(z)}$$
take $g (z)= \frac{1}{\varphi (z)} $.
