prove $f(z)=cz^n$ for some $c$. If $f$ is entire and $|f|=1$ on $|z|=1$,then $f(z)=cz^n$ for some $c$.
First consider $g(z)=f(z)/\prod(z-a_i)/(1-\overline{a_i}z)$,where $a_i$ are zeros of $f(z)$. 
Then I want to apply the maximum and minimum modulus theorem to argue that all $a_i$'s are zero. But what am I supposed to do? Do I need to first show that $g(z)$ is constant?

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*Show that $|g(z)|=1$ if $|z|=1$, and that $g$ never vanishes on the unit disk. 

*Applying maximum modulus principle to $g$ and $1/g$ we get that $g(z)=c$ on the unit disk, where $|c|=1$. 

*So we can write $f(z)=c\prod_{j=1}^n\frac{z-a_j}{1-\bar a_jz}$, and we are reduced to show that $a_j=0$ for all $j$. As $f$ is assumed entire, we need to remove the singularities in $\frac 1{\bar{a_j}}$, which has modulus $>1$. These one can't be canceled unless $a_j=0$.



Note that we need the assumption entire, otherwise, $f$ could be of the form $f(z)=c\prod_{j=1}^n\frac{z-a_j}{1-\bar a_jz}$ where $|a_j|<1$, but not necessarily equal to $0$.
