I don't know a reference, but here is a quick proof: Let the roots of the polynomial be alpha1, alpha2, ..., alphar. Let
fn(x)=prodri=1(x−alphani).
All the coefficients of fn 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 |alphai|leq1, the coefficient of xk in fn is at most binomrk.
Combining the above observations, the coefficients of the fn are integers in a range which is bounded independent of n. So, in the infinite sequence fi, only finitely many polynomials occur. In particular, there is some k and ell, with ell>0, such that f2k=f2k+ell. So raising to the 2ell power permutes the list (alpha2k1,ldots,alpha2kr). For some positive m, raising to the 2ell power m times will be the trivial permutation. In other words,
alpha2ki=alpha2k+ellmi
Every root of the above equation is 0 or a root of unity.
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