Here is a proof of alunw's claim:

Clearly all of the nonzero $\displaystyle a_j$ must have norm $\displaystyle 1$ or else $\displaystyle a_j, {a_j}^2,...$ are all distinct. Hence for these $\displaystyle a_j$ we have $\displaystyle a_j=\exp(b_j 2\pi i)$. Moreover $\displaystyle a_ja_k=\exp((b_j+b_i)2\pi i)$. Denoting the unit circle in the complex plane by $\displaystyle S_1$, define

$\displaystyle f:S_1 \rightarrow \mathbb{R}/(2\pi\mathbb{R})$

$\displaystyle \exp(b_j 2\pi i) \mapsto \overline{b_j}$

where $\displaystyle \overline{b_j}$ is the class of $\displaystyle b_j \: (\mbox{mod } 2\pi)$. Then $\displaystyle f$ is an isomorphism $\displaystyle S_1 \cong \mathbb{R}/(2\pi\mathbb{R})^+$. The nonzero $\displaystyle a_j$ are a finite subgroup $\displaystyle H$ of $\displaystyle S_1$, which maps to a finite subgroup of $\displaystyle \mathbb{R}/(2\pi\mathbb{R})^+$, which is cyclic with generator $\displaystyle \overline{2\pi/r}$ with $\displaystyle r \in \mathbb{Z}$. This implies in turn that $\displaystyle H$ is cyclic with generator $\displaystyle h=\exp( 2\pi i/r)$. Hence either $\displaystyle p(x)=\prod_{j=0}^r(x-\exp(2\pi ji/r))=x^r-1$ or $\displaystyle p(x)=x\prod_{j=0}^r(x-\exp(2\pi ji/r))=x^{r+1}-x$.