Be $f$ an holomorphic function in $D(0,1)$. Suppose that $\exists n \in \mathbb{N}$, such that $\forall r \in ]0,1[$ is verified $max \{ |f(z)| : |z| = r \} = r^n$.
Prove that $\exists \alpha \in \{ z \in \mathbb{C}$ : $|z| = 1 \}$, such that $f(z) = \alpha z^n \forall z \in D(0,1)$
I think it must be done with Maximum Modulus Principle, because it's the lesson of what it's this exercise.
Hint: Observe that $\max\{|f(z)|:|z|=r\}=r^n$ implies the following:
Now as $\mathbb{D}\setminus\{0\}$ is open connected and $\left|\frac{f(z)}{z^n}\right|$ is holomorphic on this domain, what can you conclude from the maximum modulus principle?
Consider $g(z)=\frac{f(z)}{z^n}$, then by assumption $|g(z)|\leq 1$ for $z\in D(0,1)\setminus \left\{ 0 \right\}$. Thus $z=0$ is a removable singularity of $g(z)$, we can extend $g$ to be holomorphic in $D(0,1)$. Since $|g(z)|$ attains its maximum inside $D(0,1)$, by maximum modulus principle, $g(z)=a$ with $|a|=1$ whenever $z\in D(0,1)$.
Using the finite Taylor expansion of $f$ at $z = 0$, we see that $f$ has a zero of order $n$ at $z = 0$. Let $g(z) = f(z)/z^{n-1}$, with $g(0) = 0$. By the theorem on removable singularities, $g$ is holomorphic on $D(0, 1)$. For all $z \in D(0,1)$, we have $|g(z)| \le |z|$, and for each $r \in (0, 1)$, there exists $c$ with $|c| = r$ and $|g(c)| = c$. It follows by Schwarz's lemma that $g(z) = \alpha z$ for all $z \in D(0,1)$, with $\alpha$ some constant on the unit circle. From $f(z) = z^{n-1}g(z)$ we get the desired result.
Note that the assumptions can be weakened: you can assume that max$\{|f(z)|: |z| = r\} \le r^n$ for all $r \in (0, 1)$, as long as equality holds at at least one $r$.