$f$ 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.

If $\frac{1}{f}$ has a pole of order $m$ at $z_0$, there exists a holomorphic function $g$ in a neighborhood $U$ of $z_0$ such that $\frac{1}{f(z)} = \frac{g(z)}{(z - z_0)^m}$ with $g(z_0) \neq 0$. Since $g$ is continuous and $g(z_0)\neq 0$, $g(z)\neq 0$ in a neighborhood $V$ of $z_0$. Now $U\cap V$ is a neighborhood of $z_0$ such that $f(z) = (z - z_0)^m\cdot \frac{1}{g(z)}$ for all $z\in U\cap V$, and $\frac{1}{g}$ is holomorphic on $U\cap V$ since $g$ is holomorphic and zero-free in $U\cap V$. Furthermore, $\frac{1}{g(z_0)}\neq 0$. Hence, $f$ has a zero of order $m$ at $z_0$.
