I don't know a reference, but here is a quick proof: Let the roots of the polynomial be $alpha_1$, $alpha_2$, ..., $alpha_r$. Let
$$f_n(x) = prod_{i=1}^r (x- alpha_i^n).$$
All the coefficients of $f_n$ are rational, because they are symmetric functions of the $alpha$'s, and are algebraic integers, because the $alpha$'s are, so they are integers.
Also, since $|alpha_i| leq 1$, the coefficient of $x^k$ in $f_n$ is at most $binom{r}{k}$.
Combining the above observations, the coefficients of the $f_n$ are integers in a range which is bounded independent of $n$. So, in the infinite sequence $f_i$, only finitely many polynomials occur. In particular, there is some $k$ and $ell$, with $ell>0$, such that $f_{2^k} = f_{2^{k + ell}}$. So raising to the $2^{ell}$ power permutes the list $(alpha_1^{2^{k}}, ldots, alpha_r^{2^k})$. For some positive $m$, raising to the $2^{ell}$ power $m$ times will be the trivial permutation. In other words,
$$alpha_i^{2^k} = alpha_i^{2^{k+ell m}}$$.
Every root of the above equation is $0$ or a root of unity.
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