Suppose we have an holomorphic function $f$ on the open unit disk $D(0,1)$ s.t.:
$$\forall r\in (0,1) \exists n\in \mathbb{N}| \max_{C(0,r)}|f|=r^n$$
prove that $f$ is polinomic. Honestly I don't know where to start, every approach I take leads me to triviality or useless formulas.
First note that $f$ has a zero of $n$-th order at $z=0$. Let $g(z)=\frac{f(z)}{z^n}$ and observe that $g$ attains maximum modulus in the interior.
$\textbf{EDIT:}$ I've misread the problem. We cannot draw directly from the statement the fact that there is an independent $n$ such that $\max_{|z|=r} |f(z)|=r^n$ holds. What we have is $n(r)\geq 1$ such that $$ \max_{|z|=r}|f(z)|=r^{n(r)}.$$To see there is independent $n$, note that $$ \phi(r) = \max_{|z|=r}|f(z)| $$ is continuous in $r\in(0,1)$ by the continuity of $f$. Thus, a positive integer-valued function $$ n(r) = \frac{\phi(r)}{\log r} $$ is also continuous, establishing $n(r) \equiv n$. If you feel this argument tricky, you can solve the problem inductively, by noting that if there are $r_1<r_2$ such that $n(r_1)=n(r_2)=m$ attains minimum of $n(\cdot)$, then we can argue by maximum modulus principle that $\frac{f(z)}{z^m}$ is a constant.