Let $f:D(0,1)\to D(0,1)$ holomorphic with $D(0,1)=\{ z\in\mathbb{C}: |z|<1\}$. In addition $f(0)=0$ and the root $0$ has multiplicity $N$. Show that $|f(z)|\leq |z|^N$ for all $z\in D(0,1)$. My idea: The definition of $f$ being of multiplicity $N$ yields $f(z)= z^N g(z)$ for $g(0)\neq 0$. It is obvious that ${}{}{}{}$
$$g(z)= \begin{cases} \frac{f(z)}{z^N} & z\neq 0 \\ \frac{f^{(N)}(z)}{N!} &z=0\\ \end{cases}$$
where $f^{(N)}(z)$ denotes the N-th derivative of $f$. We get $$|f(z)|=|z|^N |g(z)|$$ how do I show that $|g(z)|\leq 1$ or is this even true? Is there another way to show this inequality?