Observe that \(\abs{s}_p\lt p^{1-1/(p-1)}\) implies
\begin{align}
\abs{\zeta_\psi(1+p)^s-1}_p \amp =\abs{\zeta_\psi\exp(s\log_p(1+p))-1}_p\tag{3.25}\\
\amp =\abs{\zeta_\psi\sum_{n=0}^\infty\frac{(s\log(1+p))^n}{n!}-1}_p\tag{3.26}\\
\amp =\abs{\zeta_\psi-1+s\sum_{n=1}\frac{s^{n-1}\log_p^n(1+p)}{n!}}_p.\tag{3.27}
\end{align}
Since
\(\zeta_\psi=\psi(1+p)^{-1}\) is a root of unity of
\(p\)-power order,
\(\zeta_\psi-1\) is a uniformizer associated with a local field
\(\Q_p(\mu_{p^\nu})\) for some
\(\nu\geq1\text{.}\) Therefore, using equation
(3.27), one can easily verify that
\(|\zeta_\psi(1+p)^s-1|\lt 1\text{.}\) Therefore, the right-hand side of equation
(3.24) converges and is an analytic function of
\(s\text{.}\) Since
\(\Z\hookrightarrow\Z_p\) is dense, it suffices to derive the formula
(3.24) for
\(s=1-m\text{,}\) where
\(m\in\N\text{.}\)
We will work with \(\eta_n(\theta)\) and \(g(T,\theta)\) as \(f(T,\theta)\) is merely a normalization of \(g(T,\theta)\) and in all cases, \(\eta_n(\theta)\) and \(g(T,\theta)\) have coefficients in \(\Z_p\text{.}\) Recall the isomorphism \(\vartheta:1+p\Z_p\xrightarrow{\sim}\Z_p\)
\begin{gather*}
a\mapsto\vartheta(a)=\frac{\log_p[a]}{\log_p(1+p)}.
\end{gather*}
Therefore, \(\log_p(1+p)^{\vartheta(a)}=\log_p[a]\) and hence \(\gamma_n(a)=\gamma(1+p)^{\vartheta(a)}\text{.}\) Let \(\Z_p[T]\) be the ring of polynomials with coefficients in \(\Z_p\text{.}\) Then
\begin{gather*}
\Z_p[\Gamma_n]\cong\Z_p[T]/\ideal{(1+T)^{p^n}-1},
\end{gather*}
where the isomorphism is induced by
\(\gamma\pmod{\Gamma^{p^n}}\mapsto 1+T\pmod{(1+T)^{p^n}-1}\text{.}\) Since
\(\gamma_n(1+p)\in\Gamma/\Gamma^{p^n}\) corresponds to
\(1+T\pmod{(1+T)^{p^n}-1}\text{,}\) \(\gamma_n(a)\) corresponds to
\((1+T)^{\vartheta(a)}\pmod{(1+T)^{p^n}-1}\text{.}\) As a result, by
Proposition 3.6 and the formula
(3.20), we get that
\begin{gather}
g(T,\theta)\equiv\sum_{a\in(\Z/p^{n+1}\Z)^\times}\bparen{\frac{a(p+1)}{p^{n+1}}-\bbrace{\frac{a(1+p)}{p^{n+1}}}}\times\theta^*(a)(1+T)^{-\vartheta(a)-1}\notag\\
\pmod{(1+T)^{p^n}-1}.\tag{3.28}
\end{gather}
Applying Euclid’s Division Algorithm to \(a(1+p)\) when divided by \(p^{n+1}\text{,}\) let
\begin{gather}
a(1+p)=b_ap^{n+1}+r_a,\quad 0\leq r_a\lt p^{n+1}.\tag{3.29}
\end{gather}
Therefore,
\begin{gather*}
\frac{a(p+1)}{p^{n+1}}-\bbrace{\frac{a(1+p)}{p^{n+1}}}=b_a
\end{gather*}
for all \(a\in(\Z/p^{n+1}\Z)^\times\text{.}\) Also,
\begin{gather*}
\vartheta(a(1+p))=\frac{\log_p[a(1+p)]}{\log_p(1+p)}=\frac{\log_p[a]}{\log_p(1+p)}+1=\vartheta(a)+1
\end{gather*}
and \(a(1+p)\equiv r_a\pmod{p^{n+1}}\) implies \(\vartheta(a(1+p))\equiv\vartheta(r_a)\pmod{p^{n+1}}\text{.}\) Therefore,
\begin{gather*}
-\vartheta(a)-1\equiv-\vartheta(r_a)\pmod{p^{n}}.
\end{gather*}
Consequently, the congruence
(3.28) transforms to
\begin{gather}
g(T,\theta)\equiv\sum_{a\in(\Z/p^{n+1}\Z)^\times}b_a\theta^*(r_a)(1+T)^{-\vartheta(r_a)}\pmod{(1+T)^{p^n}-1}.\tag{3.30}
\end{gather}
For two positive integers
\(m\) and
\(n\text{,}\) where
\(n\) is sufficiently large (we need to choose
\(n\) large enough so that the character
\(\chi\) we started with may be regarded as a character of
\(\Gal(K_n/\Q)\text{,}\) in other words,
\(f_\chi\mid p^{n+1}\)). Plugging in
\(T=\zeta_\psi(1+p)^{1-m}-1\) in equation
(3.30), we get
\begin{gather}
g(\zeta_\psi(1+p)^{1-m}-1,\theta)\equiv\sum_{a\in(\Z/p^{n+1}\Z)^\times}b_a\theta^*(r_a)(\zeta_{\psi}(1+p)^{1-m})^{-\vartheta(r_a)}\notag\\
\pmod{(\zeta_\psi(1+p)^{1-m})-1}.\tag{3.31}
\end{gather}
Since \(\psi\in\widehat{\Gamma_n}\) and \(\#\widehat{\Gamma_n}=p^n\text{,}\)
\begin{gather*}
\zeta_\psi^{p^n}=\psi(1+p)^{-p^n}=1.
\end{gather*}
Therefore,
\begin{align*}
(\zeta_\psi(1+p)^{1-m})^{p^n}-1 \amp =(1+p)^{(1-m)p^n}-1\\
\amp \equiv0\pmod{p^{n+1}},
\end{align*}
which implies that
\begin{gather}
g(\zeta_\psi(1+p)^{1-m}-1,\theta)\equiv\sum_{a\in(\Z/p^{n+1}\Z)^\times}b_a\theta^*(r_a)(\zeta_{\psi}(1+p)^{1-m})^{-\vartheta(r_a)}\notag\\
\pmod{p^{n+1}}.\tag{3.32}
\end{gather}
Now, \(\log_p(1+p)^{\vartheta(r_a)}=[r_a]\) and hence \(\zeta_\psi^{-\vartheta(r_a)}=\psi(1+p)^{\vartheta(r_a)}=\psi(r_a)\text{.}\) Therefore,
\begin{gather}
g(\zeta_\psi(1+p)^{1-m}-1,\theta)\equiv\sum_{a\in(\Z/p^{n+1}\Z)^\times}b_a\theta\omega^{-1}(r_a)\psi(r_a)[r_a]^{m-1}\notag\\
\pmod{p^{n+1}}.\tag{3.33}
\end{gather}
Recall that for any \(\alpha\in\Z_p^\times\text{,}\) \(\alpha=\omega(\alpha)[\alpha]\text{.}\) Therefore,
\begin{align*}
\theta\omega^{-1}(r_a)\psi(r_a)[r_a]^{m-1} \amp =\theta\psi(r_a)\times\omega^{-m}(r_a)\times(\omega(r_a)[r_a])^{m-1}\\
\amp =\chi\omega^{-m}(r_a)r_a^{m-1}.
\end{align*}
Combining all these, we may write congruence
(3.33) as
\begin{gather}
g(\zeta_\psi(1+p)^{1-m}-1,\theta)\equiv\sum_{a\in(\Z/p^{n+1}\Z)^\times}b_a\chi\omega^{-m}(r_a)r_a^{m-1}\pmod{p^{n+1}}.\tag{3.34}
\end{gather}
We’ve chosen \(n\) large enough so that \(f_\chi\mid n\text{.}\) Thus
\begin{gather}
\chi\omega^{-m}(a(1+p))=\chi\omega^{-m}(r_a)\tag{3.35}
\end{gather}
since \(a(1+p)\equiv r_a\pmod{p^{n+1}}\text{.}\) Also,
\begin{align*}
(a(1+p))^{m} \amp =(b_ap^{n+1}+r_a)^m\\
\amp \equiv r_a^m+mr_a^{m-1}b_ap^{n+1}\pmod{p^{2(n+1)}}.
\end{align*}
Therefore,
\begin{align}
\chi\omega^{-m} \amp (1+p)(1+p)^m\sum_{a\in(\Z/p^{n+1}\Z)^\times}\chi\omega^{-m}(a)a^m\notag\\
\amp =\sum_{a\in(\Z/p^{n+1}\Z)^\times}\chi\omega^{-m}(a(1+p))(a(1+p))^m\notag\\
\amp \equiv \sum_{a\in(\Z/p^{n+1}\Z)^\times}\chi\omega^{-m}(r_a)r_a^m+mp^{n+1}\sum_{a\in(\Z/p^{n+1}\Z)^\times}b_a\chi\omega^{-m}(r_a)r_a^{m-1}\notag\\
\amp \hfill\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\pmod{p^{2(n+1)}}.\tag{3.36}
\end{align}
It is easy to see that as
\(a\) runs over
\((\Z/p^{n+1}\Z)^\times\) then so does
\(r_a\text{,}\) (recall the definition of
\(r_a\) from equation
(3.29)). As a result,
\begin{gather*}
\sum_{a\in(\Z/p^{n+1}\Z)^\times}\chi\omega^{-m}(a)a^m=\sum_{a\in(\Z/p^{n+1}\Z)^\times}\chi\omega^{-m}(r_a)r_a^m.
\end{gather*}
Since
\(\omega\) is a character modulo
\(p\text{,}\) \(\omega^{-m}(1+p)=1\text{.}\) Using congruences
(3.34) and
(3.36), we obtain,
\begin{align*}
g(\zeta_\psi(1+p)^{1-m}-1,\theta) \amp \equiv\sum_{a\in(\Z/p^{n+1}\Z)^\times}b_a\chi\omega^{-m}(r_a)r_a^{m-1}\\
\amp \equiv\frac{\bparen{(1+p)^m\chi(1+p)-1}}{mp^{n+1}}\sum_{a\in(\Z/p^{n+1}\Z)^\times}\chi\omega^{-m}(a)a^m\\
\amp \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\pmod{p^{n+1}}.
\end{align*}
\begin{align*}
g(\zeta_\psi \amp (1+p)^{1-m}-1,\theta)\\
\amp =\frac{\bparen{(1+p)^m\chi(1+p)-1}}{m}\lim_{n\to\infty}\frac{1}{p^{n+1}}\sum_{a\in(\Z/p^{n+1}\Z)^\times}\chi\omega^{-m}(a)a^m\\
\amp =-\frac{h(\zeta_\psi(1+p)^{1-m}-1)}{m}(1-\chi\omega^{-m}(p)p^{m-1})B_{n,\chi\omega^{-m}},
\end{align*}
\begin{align*}
f(\zeta_\psi(1+p)^{1-m}-1,\theta) \amp =\frac{g(\zeta_\psi(1+p)^{1-m}-1,\theta)}{h(\zeta_\psi(1+p)^{1-m}-1)}\\
\amp =-(1-\chi\omega^{-m}(p)p^{m-1})\frac{B_{n,\chi\omega^{-m}}}{m}=L_p(1-m,\chi).
\end{align*}
This completes the proof.
