Let $\mathcal O_K$ be a Dedekind domain, $L$ the fraction field of $\mathcal O_K$. Let $L/K$ be a finite, separable extension, not necessarily Galois, of degree $p$. Let $N$ be the normal closure of $L/K$. Let $G=\text{Gal}(N/K)$ and $H=\text{Gal}(N/L)$.
Let $\mathfrak p$ be a prime of $K$ (i.e. of $\mathcal O_K$). Let $Q$ be a prime of $N$ above $\mathfrak p$. Let $G_Q$ denote the decomposition group of $Q$ over $K$. Then (as discussed in Neukirch pg. 55) there is a bijection from the set of double cosets $H\backslash G/G_Q$ to the set $P_\mathfrak p$ of primes of $L$ above $\mathfrak p$, given by: $$H\backslash G/G_Q\rightarrow P_\mathfrak p, \quad H\sigma G_Q\mapsto\sigma Q\cap L$$
Now suppose the prime $\mathfrak p$ is unramified in $L$. Then $\mathfrak p$ is also unramified in $N$.
Note that there are $p$ cosets $H\sigma_1,\dots,H\sigma_p$ of $H\backslash G$ where $p=[L:K]$. There is an action of $G$ that permutes the cosets $H\sigma_i$ by right multiplication. They key observation is this:
The size of the orbit of the coset $H\sigma_i$ under the right action of the decomposition group $G_Q$ equals the inertia degree of the prime $\sigma_iQ\cap L$ over $\mathfrak p$.
To show this, first observe that, for $\rho\in G_Q$ and $\sigma_i\in G$, $$H\sigma_i\rho=H\sigma_i \Longleftrightarrow \rho\in\sigma_i^{-1}H_{\sigma_iQ}\sigma_i$$ where $H_{\sigma_iQ}$ is the decomposition group of $\sigma_iQ$ over $L$.
Thus the size of the orbit of $H\sigma_i$ is $$[G_Q:\text{stab}(H\sigma_i)]=[G_Q:\sigma_i^{-1}H_{\sigma_iQ}\sigma_i]=[\sigma_iG_Q\sigma_i^{-1}:H_{\sigma_iQ}]=[G_{\sigma_iQ}:H_{\sigma_iQ}]$$
$[G_{\sigma_iQ}:H_{\sigma_iQ}]$ equals the inertia degree of $\sigma_iQ\cap L$ over $\mathfrak p$, proving the highlighted claim above.
Now assume the degree $p$ of $L/K$ is prime, and assume that $\mathfrak p$ has two prime factors $\mathfrak P_1$ and $\mathfrak P_2$ in $L$ of degree 1. This implies we have two cosets $H\sigma_1$ and $H\sigma_2$ who orbits under the action of $G_Q$ are of size 1. $G$ is a solvable group with a transitive action on the $p$ cosets $H\sigma_1,\dots,H\sigma_p$. Thus each element of $G_Q$ fixes the two cosets $H\sigma_1$ and $H\sigma_2$, so by the theorem given in the Hint, each element of $G_Q$ must fix all the cosets, so $G_Q$ partitions the $H\sigma_i$ into $p$ distinct orbits of one element each. Thus, every prime factor of $\mathfrak p$ in $L$ is of degree 1 over $\mathfrak p$.
This is true more generally if $\zeta$ is a primitive $n$-th root of unity, where $n$ is odd (not necessarily prime).
Let $u \in \mathbb Z[\zeta]$. Let $\overline{u}$ denote the complex conjugate of $u$ (remark that this is independent of any choice of embedding of $\mathbb Q(\zeta)$ in $\mathbb C$, because $\mathbb Q(\zeta)$ is a CM field - complex conjugation always acts via $\zeta \mapsto \zeta^{-1}$).
Let $s= u/\overline{u}$.
Remark that under any embedding of $\mathbb Q(\zeta)$ in $\mathbb C$, $s$ is mapped to a complex number of absolute value $1$.
It follows that $s$ is a root of unity contained in $\mathbb Q(\zeta)$, hence $s=\zeta^a$ for some $a \in \mathbb Z/n\mathbb Z$.
Let $b \in \mathbb Z/n\mathbb Z$ be such that $2b=-a$, which exists because $n$ is odd. Let $v = \zeta^b u$. Then
$$\overline{v} = \zeta^{-b} \overline{u} = \zeta^{-b} u/s = \zeta^{-b - a} u = \zeta^b u = v.$$
Hence $v$ is totally real, hence $v \in \mathbb Z[\zeta + \zeta^{-1}]^\times$. Since $u = \zeta^{-b}v$, it follows that $\mathbb Z[\zeta]^\times = \left<\zeta\right> \mathbb Z[\zeta + \zeta^{-1}]^\times$.
Remark: This calculation is very Galois-cohomological in flavor. I'd be curious to see a proof of this result using only Galois cohomology!
Best Answer
By Kummer-Dedekind factorization theorem cited in this question, let $p$ be the minimal polynomial of $\zeta$, then if $$ p(x)\equiv\prod_ip_i(x)^{e_i}\pmod q,$$ where $p_i$ are irreducible polynomials with coefficients in $\mathbb F_q=\mathbb Z/q\mathbb Z$, then $$(q)=\prod_i(q,p_i(\zeta))^{e_i}$$ is the prime ideal factorization of $(q)$ in $\mathbb Z[\zeta]$, and the inertia degree of $(q,p_i(\zeta))$ is the degree of $p_i$. Here we know each $e_i=1$, as an aside.
Since $p(x):=x^{p-1}+\cdots+x+1=\frac{x^p-1}{x-1}$ is irreducible (by Eisenstein criterion applied to $p(x+1)$), it is the minimal polynomial for $\zeta$.
Now we try to determine the degrees of each $p_i$. By Galois theory, this is independent of $i$. So we consider only one irreducible factor $p_i$.
Suppose $\deg(p_1)=k$, and let $\theta$ be a root of $p_1$. Then $\mathbb F_q[\theta]/\mathbb F_q$ is a field extension of degree $k$. By the theory of finite fields we know that $\mathbb F_q[\theta]\cong \mathbb F_{q^k}$ and the Galois group $\operatorname{Gal}(\mathbb F_q[\theta]/\mathbb F_q)$ is generated by the Frobenius automorphism $\varphi_q:x\mapsto x^q$. That is, it sends $\theta$ to $\theta^q$. Since the order of the Galois group is just $k$, we conclude that $k$ is the smallest integer such that $\theta^{q^k}=\theta$.
But $\theta$ is a root of $p(x)$, so it has order $p$; that is, if $\theta^n=\theta$, then $p\mid n-1$. As a consequence, we see that $k$ is the smallest integer $k$ such that $p\mid q^k-1$.
Hope this helps.