Let $f\in H(D(0;R))$ and let $M(r) =\sup \{|f(z)|:|z|=r \}$ (where $r <R$). Prove that $M(r) \leq M(s)$ for $0<r<s<R$, with strict inequality if $f$ is not constant. Prove also that if $f$ is a polynomial of degree $n$ then $M(r)r^{-n}\geq M(s)s^{-n}$.
Here is what I've done so far:
Let $D(0;r)$ and $D(0;s)$. Then $f$ is holomorphic in those two sets and continuous in their boundaries since $r<s<R$. Then by the Maximum Modulus Theorem, the maximum of $|f|$ in $D(0;r)$ and $D(0;s)$ is attained on $\overline{D}(0;r)$ and $\overline{D}(0;s)$ respectively. So $\sup\{|f(z)| : |z|=r\} \leq \sup \{|f(z)| : |z|=s\}$ since $r<s$, because the circle $|z|=r$ is contained in $|z|=s$.
Now, if $f$ is constant on $D(0;R)$, $f(z)=c$ for all $z\in D(0;R)$. In particular, $M(r)=M(s)$.
For the other part, I'm stuck. But the numbers $M(r)r^{-n}$ and $M(s)s^{-n}$ are upper bounds for the coefficients of the Taylor expansion (which exists since $f \in H(D(0;R))$) of $f$ in $D(0;r)$ and $D(0;s)$ respectively, but I don't know how to exploit that fact, so I need some help with that.
Any help or hints will be appreciate, and if there are any mistakes in what I wrote for the first part, I'd be grateful if you could point them up to me. Thanks in advance!
Hint: If $f$ is a polynomial of degree $n$ then $f(z)/z^n$ has a removable singularity at $\infty$. (Or if you prefer, $z^nf(1/z)$ has a remoovable singularity at the origin.)