Why is this polynomial a monomial? Let $p$ be a polynomial of degree $n$ such that $|p(z)| = 1$ for all $|z| = 1$. 

Why is it that $p(z) = az^n$ for some $|a| = 1$?

I've noticed that we could easily prove this by induction if we could show that 0 was a root of $p$. My guess is that Rouche's theorem and/or the Maximum Modulus principle should be used.

My background to the problem: This question is the last question on the take-home portion of my final exam that I haven't been able to figure out yet. 
We are allowed to collaborate with other people in the class (and I have) as well as use any book and the internet (including this site). So basically, it's a homework assignment that is worth more points than usual. Nevertheless, in case it helps you decide how much information to give, the final is due tomorrow (but since I'm going to a math conference, I may turn it in late).

This $$\sum_{n=0}^{\infty}|a_nz^n|^2=\frac{1}{2 \pi}\int_{-\pi}^{\pi}|p(ze^{it})|^2dt$$
 kills it if you suppose that the leading coefficient of $p$ has modulus $1$.
I'm working on the general case. 
