I define the zero of order $m$ of a function $f$ at a point $z_0$ to be such that the first $m-1$ derivatives $f^{(n)}(z_0)=0$ for $n=0,\dots,m-1$ and that $f^{(m)}(z_0) \neq 0$.
I need to prove that if a holomorphic function $f$ has a zero $z_0$ of order $m$, then it can be written as $f(z)=(z-z_0)g(z)$, where $g(z)$ is holomorphic.
I know that since $f$ is holomorphic its Taylor Series expansion around $z_0$ exists for same radius of convergence.
Since $f$ has a zero of order $m$, we can write \begin{align} f(z)&=\frac{f^{(m)}(z_0)}{m!}(z-z_0)^m+\frac{f^{(m+1)}(z_0)}{(m+1)!}(z-z_0)^{m+1}+\dots\\&=(z-z_0)^m\bigg[\frac{f^{(m)}(z_0)}{m!}+\frac{f^{(m+1)}(z_0)}{(m+1)!}(z-z_0)+\dots\bigg] \end{align}
and hence calling the thing inside the parentheses $g(z)$ it seems like I'm done.
However, I am not sure why $g(z)$ should converge and if so, why it is also holomorphic.
The function $$ g_0(z)=\frac{f(z)}{(z-z_0)^m} $$ is clearly holomorphic on $\Omega\setminus\{z_0\}$ (where $\Omega$ is the domain of $f$, where it is holomorphic).
It remains to show that $g_0$ has a removable singularity at $z_0$, which follows from a repeated application of l'Hôpital or by just looking at the Taylor series expansion of $f$ at $z_0$. At the end we will find that $$ \lim_{z\to z_0}g_0(z)=\frac{f^{(m)}(z_0)}{m!}\ne0 $$
Hence we can extend $g_0$ to a function $g$ which is holomorphic on $\Omega$ and satisfies $f(z)=(z-z_0)^mg(z)$ for every $z\in\Omega$ and $g(z_0)\ne0$.
Also the converse is true: if we have $g$ holomorphic in $\Omega$, with $g(z_0)\ne0$, then $f(z)=(z-z_0)^mg(z)$ has a zero of order $m$ at $z_0$ according to your definition, by direct computation of the Taylor series at $z_0$.