Roots of unity in $\mathbb{Q}(\zeta_p)$ for $p$ an odd prime I am confused about a last step in a proof that the only roots of unity in $\mathbb{Q}(\zeta_p)$ are $\pm\zeta_p^j$, where $\zeta_p=e^{2 \pi i/p}$, $p$ is an odd prime and $1\leq j\leq p-1$.
So far it has been proven that for a root of unity $e^{2 \pi i/m} \in \mathbb{Q}(\zeta_p)$, $m$ is such that $4 \nmid p$, $q \nmid p$ where $q \neq p$ is an odd prime and $p^2 \nmid p$. This means we can only have $m|2p$. Then the proof states this implies that the only roots of unity in $\mathbb{Q}(\zeta_p)$ are $e^{2 \pi i/m} = \pm \zeta_p^j$ as required, but I don't see how to bridge this gap. Thanks for your help.

$\textbf{Note:}$ I found this proof was in fact lifted from Stewart and Tall's Algebraic Number Theory pages 189-190. They too leave this gap in the proof - they say it follows immediately but it is not obvious to me at all.
$\textbf{Further edit:}$ In Stewart and Tall, the condition on $m$ is that $m|2p$, not $m = \pm 2p$ as I said. I have changed this above.
 A: We know that any root of unity $z$ can be expressed as $z= e^{2\pi i k/m}$ where $k$ and $m$ are relatively prime ($z$ is a primitive $m$th root of unity for some $m$).  Since $(k,m)=1$, there is a multiplicative inverse of $k$ mod $m$, call it $l$.  If $z \in \mathbb Q(\zeta_p)$, then since fields are closed under multiplication, $z^l \in \mathbb Q(\zeta_p)$, and $z^l = e^{2\pi i kl/m} = e^{2\pi i/m}$.
By the facts mentioned in the question, $m \mid 2p$, which means that any root of unity in $\mathbb Q(\zeta_p)$ must be a $2p$-th root of unity.  The $2p$-th roots of unity can all be written as one of $\zeta_p^j$ or $-\zeta_p^j$ for some $j$.  To see this note that when $k = 2j$ is even, $e^{2\pi i k/2p} = \zeta_p^j$, while if $k=2j+1$ is odd, $e^{2\pi i k/2p} = -\zeta_p^{j+(p+1)/2}$.
A: You have proven that $m=\pm2p$. So  

$$(e^{\frac{2\pi i}{m}})^2=e^{\frac{\pm2\pi i}{p}}=e^{\pm\frac{2\pi i(p+1)}{p}}.$$
  Hence $$e^{\frac{2\pi i}{m}}=\pm e^{\frac{2\pi ik}{p}},$$
  where $k=\pm \frac{(p+1)}{2}$   

Barring mistakes. Thanks in advance.  
