Using the Euler product
(2.11) of Dedekind zeta functions, we can write
\(\zeta_K(s)\) as a product over rational primes as follows
\begin{gather}
\zeta_K(s)=\prod_{\mf{p}}\bparen{1-\frac{1}{N(\mf{p})^s}}^{-1}=\prod_{p}\prod_{\mf{p}|p}\bparen{1-\frac{1}{N(\mf{p})^s}}^{-1}.\tag{2.19}
\end{gather}
In the second product, the inner product runs over all prime ideals
\(\mf{p}\) over
\(p\text{,}\) and we denote that using the standard symbol
\(\mf{p}|p\text{.}\) On the other hand, the left-hand side of equation
(2.18) can be written as a product over integer primes as well, using the Euler product
(2.7) of Dirichlet
\(L\)–functions:
\begin{gather}
\prod_{\chi\in H}L(s,\chi)=\prod_{p}\prod_{\chi\in H}\bparen{1-\frac{\chi(p)}{p^s}}^{-1}.\tag{2.20}
\end{gather}
\begin{gather}
\prod_{\mf{p}|p}\bparen{1-\frac{1}{N(\mf{p})^s}}^{-1}= \prod_{\chi\in H}\bparen{1-\frac{\chi(p)}{p^s}}^{-1}.\tag{2.21}
\end{gather}
for each rational prime
\(p\text{.}\) Note that
\(\Gal(\Q(\mu_n)/\Q)\) is abelian; hence, any subgroup is normal. Hence, by the Fundamental Theorem of Galois Theory
[28, Chapter VI, Theorem 1.1],
\(K/\Q\) is a Galois extension.
Fix an integer prime \(p\) and Let \(e_p\) and \(f_p\) denote the ramification index and inertia degree respectively and \(g_p\) denote the number primes in \(\mathcal{O}_K\) lying over \(p\text{.}\) Then,
\begin{align}
\prod_{\mf{p}|p}\bparen{1-\frac{1}{N(\mf{p})^s}}^{-1} \amp =\bparen{1-\frac{1}{p^{f_ps}}}^{-g_p}\tag{2.22}
\end{align}
for each integer prime
\(p\text{.}\) First, suppose that
\(p\nmid n\text{.}\) Then
\(\Q(\mu_n)\) and, in particular,
\(K\) is unramified over
\(\Q\) at
\(p\text{.}\) Consequently,
\(e_p=1\) and hence
\([K:\Q]=f_pg_p\text{.}\) Since
\(K/\Q\) is unramified at
\(p\) and
\(\Gal(K/\Q)\) is a subgroup of
\((\Z/n\Z)^\times\text{;}\) hence abelian, the Frobenius map
\(x\mapsto x^p\) in
\(\Gal(\mathcal{O}_K/\mf{p}/\F_p)\text{,}\) which is of order
\(f_p\text{,}\) lifts to an order
\(f_p\) element, denoted as
\(\Frob{p}\text{,}\) in
\(\Gal(K/\Q)\text{.}\) Regard
\(H\) as the group of characters of
\(\Gal(K/\Q)\text{.}\) For any
\(\chi\in H\text{,}\) note that
\(\chi(\Frob{p})\) is a complex
\(f_p^{\rm th}\) root of unity. Consider the map
\(\Phi:H\longrightarrow \mu_{f_p}\) given by
\(\chi\mapsto\chi(\Frob{p})\text{,}\) where we think of
\(\mu_{f_p}\) as a multiplicative subgroup of
\(\C^\times\text{.}\) Suppose that
\(\chi\in\ker(\Phi)\text{,}\) i.e., \(\chi(\Frob{p})=1\text{.}\) Then
\(\chi\) can be considered as a character on
\(\Gal(K/\Q)/\ideal{\Frob{p}}\text{,}\) where
\(\ideal{\Frob{p}}\) is the cyclic group of order
\(f_p\) generated by
\(\Frob{p}\text{.}\) Conversely, any character
\(\chi\) of
\(\Gal(K/\Q)/\ideal{\Frob{p}}\) can be extended to a character
\(\chi\in H\) such that
\(\chi(\Frob{p})=1\text{.}\) Therefore,
\(\#\ker(\Phi)=\#\Gal(K/\Q)/f_p=g_p\text{.}\) By the First Isomorphism Theorem of groups
[28, Chapter I, p. §3],
\(\Im(\Phi)\subset\mu_{f_p}\) has
\(f_p=\#\mu_{f_p}\) elements; hence
\(\Im(\Phi)=\mu_{f_p}\text{.}\) This shows that
\(\Phi\) is surjective and as
\(\chi\) varies over
\(H\text{,}\) \(\chi(p)=\chi(\Frob{p})\) varies over all the complex
\(f_p^{\rm th}\) roots of unity, where each of them occur exactly
\(g_p\) times (via the isomorphism
\((\Z/n\Z)^\times\cong\Gal(\Q(\mu_n)/\Q)\text{,}\) \(p\pmod{n}\in(\Z/n\Z)^\times\) corresponds to
\(\Frob{p}\in\Gal(K/\Q)\)). Therefore,
\begin{gather}
\prod_{\chi\in H}\bparen{1-\frac{\chi(p)}{p^s}}^{-1}=\prod_{\zeta^{f_p}=1}\bparen{1-\frac{\zeta}{p^s}}^{-g_p}.\tag{2.23}
\end{gather}
Now, recall that
\(\prod_{\zeta^{f_p}=1}(1-\zeta X)=1-X^{f_p}\) in
\(\C[X]\text{.}\) Therefore, equation
(2.23) transforms to
\begin{gather}
\prod_{\chi\in H}\bparen{1-\frac{\chi(p)}{p^s}}^{-1}=\bparen{1-\frac{1}{p^{f_ps}}}^{-g_p}.\tag{2.24}
\end{gather}
Comparing equations
(2.22) and
(2.24), we conclude that equation
(2.20) holds in the case
\(p\nmid n\text{.}\)
Now, consider the case when
\(p\mid n\text{.}\) Following the notations used in
Lemma 2.17, let
\(K'=K\cap\Q(\mu_{n'})\text{.}\) Then, by
Lemma 2.17,
\(K'\) is the maximal extension of
\(\Q\) in
\(K\) that is unramified at
\(p\text{.}\) Let
\(\mf{p}\) be a prime in
\(K'\) lying above
\(p\) and
\(\mf{P}\) be a prime in
\(K\) lying above
\(\mf{p}\text{.}\) Then we have a tower of extensions
\(K_{\mf{P}}/K'_{\mf{p}}/\Q_p\) local fields, with
\([K_{\mf{P}}/\Q_p]=e_pf_p\text{.}\) Now,
\(K'_{\mf{p}}/\Q_p\) is the unique maximal unramified extension in
\(K_{\mf{P}}\) corresponding to the finite field extension
\(\mathcal{O}_{K'}/\mf{p}/\F_p\) and hence
\([K'_{p}:\Q_p]=f_p\text{.}\) Therefore,
\([K_{\mf{P}}:K_{\mf{p}}]=e_p\text{,}\) which implies that
\(K_{\mf{P}}/K'_{\mf{p}}\) is totally ramified. This is true for all primes
\(\mf{p}\) of
\(K'\) lying above
\(p\text{.}\) As a result, if we pass from
\(K\) to
\(K'\text{,}\) the product in equation
(2.22) does not change as
\(f_p\) and
\(g_p\) does not change. Moreover, whenever
\(p\) divides the conductor of
\(\chi\in H\text{,}\) \(\chi(p)=0\) and the Euler factor at
\(p\) in equation
(2.23) does not contribute anything. So, we pass to the subgroup
\(H'\) of
\(H\) of characters whose conductors are relatively prime to
\(p\text{.}\) Therefore,
\begin{gather}
\prod_{\chi\in H}\bparen{1-\frac{\chi(p)}{p^s}}^{-1}=\prod_{\chi\in H'}\bparen{1-\frac{\chi(p)}{p^s}}^{-1}.\tag{2.25}
\end{gather}
Therefore, we are left to prove that
\(K'=\Q(\mu_{n'})^{H'^{\perp}}\text{.}\) Note that
\(H'=H\cap\mathcal{X}(n')\) by
Lemma 2.17. By
[47, Corollary 3.6],
\(K'=\Q(\mu_n)^{H'^\perp}\text{.}\) Now, by
Lemma 2.17,
\begin{gather}
K'=K'\cap\Q(\mu_{n'})=\Q(\mu_{n'})\cap\Q(\mu_n)^{H'\perp}=\Q(\mu_{n'})^{H'^\perp}.\tag{2.26}
\end{gather}
Now, since passing from
\(K\) to
\(K'\) does not change the product
(2.22), passing from
\(H\) to
\(H'\) does not change the product on the right-hand side of equation
(2.25), and
\(K'=\Q(\mu_{n'})^{H'^\perp}\text{,}\) our argument in the first part of the proof applies as
\(p\nmid n'\text{.}\) This completes the proof.