Schwarz's Lemma exercise
Define a function $g$ on $D(0,1)$ by $g(0)=0$ and $$ g(z) = \frac{f(z)}{z^{n-1}}, z \in D(0,1) \setminus \{0\}.$$
Then $g$ is holomorphic in $D(0,1)$ and $g(0)=0$. Thus, by schwarz lemma, $|g(z)| \leq |z|$ for all $z \in D(0,1)$, that is, $|f(z)| \leq |z|^n$ for all $z \in D(0,1) \setminus \{0\}.$
Moreover, since $f(0)=0$, then $|f(z)| \leq |z|^n$ for all $z \in D(0,1).$