My main question is the title: for an odd prime $p$, denote a primitive $p^{\text{th}}$ root of unity by $\zeta_p$. Is it true that $i$ is not contained in the cyclotomic extension $\mathbb{Q}(\zeta_p)$? If this is true, is the following proof correct, and if it is not true, where does the following proof break down:
Recall that the unique quadratic subfield of $\mathbb{Q}(\zeta_p)$ is $\mathbb{Q}(\sqrt{\pm p})$ where there is a $"+"$ is $p\equiv 1 \mod 4$ and a $"-"$ if $p\equiv 3 \mod 4$ (Source: exercise 11 of section 14.7 of Dummit and Foote). Assume to the contrary that $i \in \mathbb{Q}(\zeta_p)$. Then $\mathbb{Q}(i)$ is a quadratic extension of $\mathbb{Q}$ of degree 2 contained in $\mathbb{Q}(\zeta_p)$. But this yields an immediate contradiction since of course $\mathbb{Q}(\sqrt{\pm p}) \neq \mathbb{Q}(i)$. So $i\notin \mathbb{Q}(\zeta_p)$.