Hermite–Lindemann gives all-integer phase independence

Assume the proposition that, for every complex number $\beta$, if $\beta$ is algebraic over $\mathbb Q$ and nonzero, then $\exp(\beta)$ is transcendental over $\mathbb Q$. For every complex number $\alpha$ that is algebraic over $\mathbb Q$ and satisfies $\alpha\ne0$, the integer-indexed family $\left(\exp\!\left((n:\mathbb C)(\mathrm i\alpha)\right)\right)_{n\in\mathbb Z}$ is linearly independent over the algebraic closure of $\mathbb Q$ inside $\mathbb C$. Explicitly, every finitely supported family of coefficients $a_n$ from that algebraic closure which satisfies $\sum_{n\in\mathbb Z}a_n\exp((n:\mathbb C)(\mathrm i\alpha))=0$ has $a_n=0$ for every $n$. The index set is all of $\mathbb Z$, including $0$ and negative integers; the $n=0$ member is $\exp(0)$, while negative indices use negative integer multiples in the exponential. No assumption says that $\alpha$ is real. The assertion is conditional on supplied proofs of the global exponential-transcendence proposition, algebraicity of $\alpha$, and $\alpha\ne0$; when any required hypothesis is unavailable, the theorem supplies no conclusion for that case.