Exercise 8.26: A multiplicity proof

Suppose that $f$ is holomorphic at $z_0$, $f(z_0) = 0$, and $f'(z_0)\neq0$. Show that $f$ has a zero of multiplicity $1$ at $z_0$.

Here's my attempt:

Let $f$ be holomorphic at $z_0$, $f(z_0) = 0$, and $f'(z_0) \neq 0$. Thus, we can write $f(z_0) = (z-z_0)^mg(z)$ where $g(z)$ is holomorphic and nonzero at $z_0$, and $m \in \mathbb{N}$

If the derivative of $f$ is nonzero, this explicitly means
$$
\lim_{z\rightarrow z_0} \frac{f(z+z_0) - f(z_0)}{z-z_0} \neq 0,
$$

but $f(z_0) = 0$, so
$$
\lim_{z\rightarrow z_0} \frac{f(z+z_0)}{z-z_0}= \lim_{z\rightarrow z_0} \frac{(z-z_0)^m g(z)}{z-z_0} = \lim_{z\rightarrow z_0} (z-z_0)^{m-1} g(z) \neq 0.
$$

Since $g(z)$ must be continuous and $g(z_0) \neq 0$, this can only be true if $m=1$.

Yes, I think this works. I think it might be simpler if you work directly with the series representations.