I want to show that a number field of odd degree contains only $2$ roots of unity. The only information I really have regarding this that I think is relevant is that the group of units $\mathcal{O}_K^\times$ of a number field $K$ is such that $$\mathcal{O}_K^\times \cong \mathbb{Z}^{r_1+r_2-1} \times G$$ where $(r_1,r_2)$ is the signature of K, and $G$ is the finite cyclic group consisting of the roots of unity that lie in $K$. Then I am reduced to showing that $G$ only consists of two elements.
I know that the degree of a number field $n=r_1+2r_2$, so an odd degree $n$ means that that value of $r_1$ must be odd also.
My problem is, I am not sure what that shows - or even if it is relevant! The only consequence of this then is that the parity of $r_1+r_2-1$ reflects the parity of $r_2$, and I am not really sure if that means anything for the size of $G$. However, I cannot think of any other information given that suggests anything in terms of $\mathcal{O}_K^\times$, $r_1+r_2-1$ or $G$.
Hint: $[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\varphi(n)$ is even for all $n\geq 3$.