If $p$ is a polynomial, then either $\,p(z)=z^n\,$ or $\,\max_{|z|=1}|p(z)|>1$. Let $p(z)=z^{n}+a_{n-1}z^{n-1}+...+a_{0}$ be a polynomial of degree $n\geq1$.
How can you prove that either $p(z)=z^{n}$ or there exists $z'$ with $|z'|=1$ such that $|p(z')|>1$.
Maybe we can use the maximum modulus principle and consider $q(z)=z^{n}p(1/z)$.
 A: Let us assume that $\lvert f(z)\rvert\le 1$, for all $\lvert z\rvert=1$, and show that in such case $f$ has to be equal to $z^n$.
Cauchy Integral formula implies that
$$
n!=f^{(n)}(0)=\frac{n!}{2\pi i}\int_{|z|=1}\frac{f(z)\,dz}{z^{n+1}}=\frac{n!}{2\pi }\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt,
$$
and thus, $\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt\in\mathbb R\,$ and
$$
1=\frac{1}{2\pi }\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt\le 
\frac{1}{2\pi }\int_0^{2\pi} |f(\mathrm{e}^{it})|\,dt\le 1,
\tag{1}
$$
since $\lvert\, f(z)\rvert\le 1$, for all $\lvert z\rvert=1$. Therefore,
$$
\mathrm{Re}\,\big(f(\mathrm{e}^{it})\,
\mathrm{e}^{-int}\big)=|f(\mathrm{e}^{it})|=1, 
$$
for all $\,t\in[0,2\pi]$, and thus
$$\,\mathrm{Im}\,\big(f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\big)=0,$$
for all $\,t\in[0,2\pi]$. Consequently,
$$
f(\mathrm{e}^{it})\,\mathrm{e}^{-int}=1,\quad\text{for all $t\in[0,2\pi]$,}
$$  or equivalently
$$
f(z)=z^n,
$$
for $\lvert z\rvert=1$, which implies that $f(z)=z^n$.
Extension. From the proof above we can make the following generalization:
Let $f:U\to\mathbb C$ be analytic, where $U$ contains the closed unit disk. If $f^{(n)}(0)=n!$
then
$$
\max_{|z|=1}|f(z)|>1,
$$
unless $f(z)=z^n$.
