Notations
We denote by $|S|$ the number of elements of a finite set $S$.
Let $A$ be a ring.
We denote by $A^{\times}$ the group of invertible elements of $A$.
Let $f(X) \in \mathbb{Z}[X]$ and $p$ be a prime number.
We denote by $\bar f(X)$ the reduction of $f(X)$ (mod $p$).
Fixed symbols
We fix the following symbols.
Let $l$ be a prime number(possibly 2).
Let $h \geq 1$ be an integer.
Let $\zeta$ be a primitive $l^h$-th root of unity in $\mathbb{C}$.
Let $K = \mathbb{Q}(\zeta)$.
Lemma 1
Let $\Omega$ be an algebraically closed field of characteristic $p$($p$ may be 0).
Suppose $p \neq l$.
Let $S$ be the set of roots of $X^{l^h} - 1$ in $\Omega$.
Then $S$ is a cyclic subgroup of $\Omega^{\times}$ of order $l^h$.
Let $P$ be the set of generators of $S$.
Then $|P| = l^{h-1}(l - 1)$.
Let $\Phi(X) = \prod_{\omega\in P}(X - \omega)$.
Then $\Phi(X) = (x^{l^{h-1}})^{l-1} + (x^{l^{h-1}})^{l-2} +\cdots+ x^{l^{h-1}} + 1$.
Proof:
Since $(X^{l^h} - 1)' = l^hX^{l^h-1}$ and $l \neq p$, $X^{l^h} - 1$ has no multiple roots in $\Omega$. Hence $|S| = l^h$.
It is well known that $S$ is a cyclic subgroup of $\Omega^{\times}$.
Let $P$ be the set of generators of $S$.
Then $|P| = l^h - l^{h-1} = l^{h-1}(l - 1)$.
$\Phi(X) = (X^{l^h} - 1)/(X^{l^{h-1}} - 1) = (X^{l^{h-1}})^{l-1} + (X^{l^{h-1}})^{l-2} +\cdots+ X^{l^{h-1}} + 1$.
QED
Definition
$\Phi(X)$ of Lemma 1 is called the cyclotomic polynomial of order $l^h$.
Lemma 2
Let $\zeta', \zeta$ be primitive $l^h$-th roots of unity in $\mathbb{C}$.
Then $(1 - \zeta')/(1 - z)$ is a unit of $K$.
Proof:
There exists an integer $a \geq 1$ such that $\zeta' = \zeta^a$.
Hence $(1 - \zeta')/(1 - \zeta) = \zeta^{a-1} +\cdots+ \zeta + 1$.
Hence $(1 - \zeta')/(1 - \zeta)$ is an algebraic integer in $K$.
Similarly $(1 - \zeta)/(1 - \zeta')$ is an algebraic integer in $K$.
Hence $(1 - \zeta')/(1 - \zeta)$ is a unit of $K$.
QED
Lemma 3
The following assertions hold.
(1) The cyclotomic polynomial $\Phi(X)$ of order $l^h$ is irreducible in $\mathbb{Q}[X]$.
(2) $\mathfrak{l} = (1 - \zeta)$ is a prime ideal of degree 1 of $K$.
(3) $l = \mathfrak{l}^{l^{h-1}(l - 1)}$.
Proof:
Let $P$ be the set of primitive $l^h$-th roots of unity in $\mathbb{C}$.
By Lemma 1, $\Phi(X) = \prod_{\zeta\in P}(X - \zeta) = (X^{l^{h-1}})^{l-1} + (X^{l^{h-1}})^{l-2} +\cdots+ X^{l^{h-1}} + 1$.
Substituting $X = 1$, we get $l = \prod_{\zeta\in P}(1 - \zeta)$. By Lemma 2, $(l) = \mathfrak{l}^{l^{h-1}(l - 1)}$, where $\mathfrak{l} = (1 - \zeta)$.
This proves (3).
Hence $l^{h-1}(l - 1) \leq [K : \mathbb{Q}]$.
On the other hand, since $\Phi(\zeta) = 0$, $[K : \mathbb{Q}] \leq l^{h-1}(l - 1)$.
Hence $l^{h-1}(l - 1) = [K : \mathbb{Q}]$.
Hence $\Phi(X)$ is irreducible.
This proves (1).
Since $[K : \mathbb{Q}] = l^{h-1}(l - 1)$, $\mathfrak{l}$ is a prime ideal of degree 1 by (3). This proves (2).
QED
Lemma 4
The notations are the same as Lemma 3.
$(\Phi'(\zeta)) = \mathfrak{l}^{l^{h-1}(h(l - 1) - 1)}$
Proof:
$X^{l^h} - 1 = \Phi(X)(X^{l^{h-1}} - 1)$.
Taking derivatives of the bothe sides, we get
$l^hX^{l^h-1} = \Phi'(X)(X^{l^{h-1}} - 1) + l^{h-1}\Phi(X)X^{l^{h-1}-1}$.
Substituting $X = \zeta$, we get $l^h\zeta^{l^h-1} = \Phi'(\zeta)(\zeta^{l^{h-1}} - 1)$.
Since $\zeta^{l^{h-1}}$ is a primitive $l$-th root of unity,
$(l) = (\zeta^{l^{h-1}} - 1)^{l-1}$ in $\mathbb{Q}(\zeta^{l^{h-1}})$.
Since $(l) = \mathfrak{l}^{l^{h-1}(l - 1)}$ in $K$,
$(\zeta^{l^{h-1}} - 1) = \mathfrak{l}^{l^{h-1}}$ in $K$.
Hence $\mathfrak{l}^{hl^{h-1}(l - 1)} = \Phi'(\zeta)\mathfrak{l}^{l^{h-1}}$.
Hence $(\Phi'(\zeta)) = \mathfrak{l}^{l^{h-1}(h(l - 1) - 1)}$.
QED
Lemma 5
The notations are the same as Lemma 3.
Let $d$ be the discriminant of $\Phi(X)$.
Then $|d| = l^{l^{h-1}(h(l - 1) - 1)}$.
Proof:
By Lemma 4, $(\Phi'(\zeta)) = \mathfrak{l}^{l^{h-1}(h(l - 1) - 1)}$.
Taking the norms of the both sides, we get $|d| = |N_{K/\mathbb{Q}}(\Phi'(\zeta))| = N(\mathfrak{l})^{l^{h-1}(h(l - 1) - 1)}$.
Since $N(\mathfrak{l}) = l$, $|d| = l^{l^{h-1}(h(l - 1) - 1)}$.
QED
Lemma 6
$\mathbb{Z}[\zeta]$ is the ring of algebraic integers of $K$.
Proof:
This follows from Lemma 3, Lemma 5 and the proposition of my answer to this question.
Lemma 7
Let $\Phi(X)$ be the cyclotomic polynomial of order $l^h$ in $\mathbb{Q}[X]$.
Let $p$ be a prime number such that $p \neq l$.
Let $X^{l^h} - 1 \in \mathbb{Z}[X]$.
Since $(X^{l^h} - 1)' = l^hX^{l^h-1}$, $X^{l^h} - 1$ has no multiple irreducible factor mod $p$.
Since $X^{l^h} - 1 = \Phi(X)(X^{l^{h-1}} - 1)$,
$\Phi(X)$ has no multiple irreducible factor mod $p$.
Let $\Phi(X) \equiv f_1(X)\cdots f_r(X)$ (mod $p$), where $f_i(X)$ are distinct monic irreducible polynomials mod $p$.
Let $f$ be the smallest positive integer such that $p^f \equiv 1$ (mod $l^h$).
Then the degree of each $f_i(X)$ is $f$.
Proof:
Let $F = \mathbb{Z}/p\mathbb{Z}$.
Let $\Omega$ be the algebraic closure of $F$.
Let $S$ be the set of roots of $X^{l^h} - 1$ in $\Omega$.
Since $(X^{l^h} - 1)' = l^hX^{l^h-1}$, $X^{l^h} - 1$ has no multiple roots in $\Omega$.
Hence $|S| = l^h$.
It is well known that $S$ is a cyclic subgroup of $\Omega^{\times}$.
Let $P$ be the set of generators of $S$.
$|P| = l^h - l^{h-1} = l^{h-1}(l - 1)$.
Let $\bar \Phi(X) = \prod_{\omega\in P}(X - \omega) \in F[X]$.
Then $\bar \Phi(X) = (X^{l^{h-1}})^{l-1} + (X^{l^{h-1}})^{l-2} +\cdots+ X^{l^{h-1}} + 1$.
Hence $\bar \Phi(X) = \bar f_1(X)\cdots\bar f_r(X)$.
Let $\omega$ be a root of \bar f_i(X) in $\Omega$.
Since $\omega$ is a root of $\bar \Phi(X)$, $\omega \in P$.
Let $E$ be the unique subfield of $\Omega$ such that $|E| = p^f$.
It is well known that $E^{\times}$ is a cyclic group.
Since $|E^{\times}| = p^f - 1$ and $l^h|p^f - 1$, $E^{\times}$ has a unique cyclic subgroup of order $l^h$.
Hence $\omega \in E$.
Let $L$ be a proper subfiled of $E$.
Let $[L : F] = p^r$.
Suppose $\omega \in L$.
Then $l^h|p^r - 1$, i.e. $p^r \equiv 1$ (mod $l^h$).
Since $r < f$, this is a contradiction.
Hence $E = F(\omega)$.
Hence the minimal polynomial of $\omega$ over $F$ has degree $f$.
Since $\bar f_i(X)$ is irreducible, $\bar f_i(X)$ is the minimal polynomial of $\omega$.
This completes the proof.
QED
Proposition
The only prime number which ramifies in $K$ is $l$, except $l = 2$ and $h = 1$.
Proof:
By Lemma 3, $l$ ramifies in $K$ except $l = 2$ and $h = 1$.
Let $p$ be a prime number such that $p \neq l$.
Let $\Phi(X) \equiv f_1(X)\cdots f_r(X)$ (mod $p$) be as in Lemma 7.
By this question, $P_i = (p, f_i(\zeta))$ is a prime ideal of $\mathbb{Z}[\zeta]$ lying over $p\mathbb{Z}$.
It is easy to see that $\mathbb{Z}[\zeta]/P_i$ is a finite extension of $\mathbb{Z}/p\mathbb{Z}$ of degree $f$.
By Lemma 6, $\mathbb{Z}[\zeta]$ is the ring of algebraic integers in $K$.
It is well known that each $P_i$ has the same ramification index $e$ and $l^{h-1}(l - 1) = efg$, where $g$ is the number of prime ideals of $\mathbb{Z}[\zeta]$ lying over $p\mathbb{Z}$.
Since $l^{h-1}(l - 1) = fr$, $r = g$ and $e = 1$.
Hence $p$ does not ramify in $K$.
This completes the proof.
QED
Best Answer
The OP has already suggested two ways to solve this:
(a) Let $\zeta_n$ generated the group of roots of unity in $\mathbb Q(\zeta_l)$. Then $2l$ divides $n$, and also $\mathbb Q(\zeta_l) = \mathbb Q(\zeta_n)$. A consideration of degrees shows that $\varphi(n) = \varphi(l)$, and combining this with the fact that $2l$ divides $n$, elementary number theory implies that in fact $n = 2l$.
(b) Ramification theory rules out the possibility of $\zeta_n$ lying in $\mathbb Q(\zeta_l)$ if $n$ is divisible by an odd prime $p \neq l$ or by a power of $2$ greater than the first.
Here are some other arguments (I continue to let $\zeta_n$ be the generator of the roots of unity in $\mathbb Q(\zeta_l)$):
(c) Galois theoretic: since $\mathbb Q(\zeta_l) = \mathbb Q(\zeta_n)$, passing to Galois groups over $\mathbb Q$, we find that the units in $\mathbb Z/n$ project isomorphically onto the units in $\mathbb Z/l$. Given that $2l | n$, we deduce from the Chinese remainder theorem that $n = 2l$.
(d) Discriminants: Since $\mathbb Q(\zeta_l) = \mathbb Q(\zeta_n)$, a consideration of the standard discriminant formulas shows that $n = 2l$.
(e) Looking at the reduction modulo split primes: Choose $p$ prime to $n$ and congruent to $1$ mod $l$. Then the group of $n$th roots of unity injects into the residue field of any prime lying over $p$. Since $p \equiv 1 \bmod l$, this residue field is just $\mathbb F_p$, and so we find that $n | p-1$ if $p > n$ (say) and $p \equiv 1 \bmod l$. Dirichlet's theorem then gives that the units in $\mathbb Z/n$ project isomorphically onto the units in $\mathbb Z/l$, from which we deduce that $n = 2l$.
(f) Working locally at l: it is not hard to check that the roots of unity in $\mathbb Q_l(\zeta_l)$ are precisely $\mu_{l(l-1)}$. So we have to show that the only $(l-1)$st roots of $1$ in $\mathbb Q(\zeta_l)$ are $\pm 1$. Actually I don't see how to do this right now without reverting to one of the other arguments, but there's probably a pithy way.
Note that (c) is just a fancy version of (a), while (d) is a more concrete form of (b) (which uses less theory). It may seem that (e) is overkill, and it certainly is for this question, but the method can be useful, and it has an obvious connection to (c) via reciprocity laws. Method (f) (unfortunately incomplete) is related to (b).