On a proof of a number field theorem 
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*$\textbf{Theorem}\ $For every integer $n\geq 2$, let $\mathbb{Q}[\zeta_n]$ be the cyclotomic field generated by the primitive $n$-th roots of $1$. Suppose $\mathbb{Q}[\zeta_m]=\mathbb{Q}[\zeta_n]$ with $m<n$. Then $m$ is odd and $n=2m$.


What I know:


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*the degree $[\mathbb{Q}[\zeta_n]:\mathbb{Q}]=\phi(n)$;

*the discriminant $d_n=\operatorname{disc}(\zeta_n)$ divides $n^{\phi(n)}$;

*a rational prime $p$ ramifies in the ring of integers of $\mathbb{Q}[\zeta_n]$ if and only if $p$ divides $d_n$.
What I know is sufficient to prove the Theorem? If yes, how can I proceed?
 A: Here is one-possible approach:
If $\mathbb{Q}(\zeta_m)=\mathbb{Q}(\zeta_n)$ by comparing degrees we get that $\varphi(m)=\varphi(n)$.
But, it's easy to see that $\mathbb{Q}(\zeta_n)\cap\mathbb{Q}(\zeta_m)=\mathbb{Q}(\zeta_{(n,m)})$, so we can actually assume WLOG that $n\mid m$.
But, factor $m=p_1^{e_1}\cdots p_\ell^{e_\ell}$ so then by assumption, with the possibility of reordering, $n=p_1^{f_1}\cdots p_k^{f_k}$ with $k\leqslant\ell$ and $f_j\leqslant e_j$. So, then, 
$$\varphi(m)=p_1^{e_1-1}(p_1-1)\cdots p_\ell^{e_\ell-1}(p_\ell-1)$$
and
$$\varphi(n)=p_1^{f_1-1}(p_1-1)\cdots p_k^{f_k-1}(p_k-1)$$
By assumption, $\varphi(m)=\varphi(n)$. But, upon division we find that this implies that 
$$p_1^{e_1-f_1}\cdots p_k^{e_k-f_k}p_{k+1}^{e_{k+1}-1}(p_k-1)\cdots p^{e_\ell-1}(p_\ell-1)=1$$
Clearly we must have that $e_i-f_i=0$ for $i=1,\ldots,k$ and $e_i-1=0$ for $i=k+1,\ldots,\ell$ and also that $p_i-1=1$ for $i=k+1,\ldots,\ell$. Thus, if $k\ne\ell$ This last condition automatically implies that $p_i=1$ for $i=k+1,\ldots,\ell$ and since the $p_i$ are distinct this actually implies that $\ell=k+1$ and $p_\ell=2$. So then, it's easy to see that 
$$m=2 p_1^{e_1}\cdots p_k^{e_k}$$
and 
$$n=p_1^{e_1}\cdots p_k^{e_k}$$
which is exactly what you wanted.
