If $|p(z)|\leq1$ for all $|z|\leq1$ then $p(z)=z^n$

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I have a polynomial $p(z)=z^n+a_{n-1}z^{n-1}+..+a_0$ satisfying $|p(z)|\leq 1$ for all $|z|\leq 1$. I have to show that $a_{n-1}=...=a_0=0$.
Applying Cauchy estimates gave me $|a_k|\leq 1$ for all $k$, which doesn't seem to help. For non-zero $z$ I can write $|\frac{z^n}{|z|^n}+..+a_0|\leq 1$ but I don't see how that would help either. Any hint would be appreciated.