Schwarz Lemma - like exercise
Solution 1:
For $0< r<1$, let $D_r=\{z\in\mathbb{C}:|z|\le r\}$.
The function $g(z)=\dfrac{f(z)}{z^m}$ is analytic on $D$ (see Removable Singularity) and $|g(z)|\le\frac{1}{r^m}$ on $\partial D_r$. The maximum modulus principle says that $$ |g(z)|\le\frac{1}{r^m}\text{ for }z\in D_r\tag{1} $$ Since $(1)$ holds for all $r<1$, we have that $|g(z)|\le1$ for $z\in D$, and therefore, $$ |f(z)|\le|z^m|\tag{2} $$