Write
\(\chi_i=\theta_i\psi_i\) where
\(\theta_i\in\widehat{\Delta}\) and
\(\psi_i\in\widehat{\Gamma_m}\text{.}\) Consider the characters
\(\chi_{i,n}\coloneqq\chi_i\omega^n\text{.}\) From the condition that
\(\chi_i\) do not contain
\(\omega^{-n}\) modulo
\(p\text{,}\) we conclude that
\(\theta_i\omega^n\neq1\) and hence, by
Lemma 4.10, we get that
\(f(0,\theta_1\omega^n)\equiv f(0,\theta_2\omega^n)\pmod{\mf{p}}\text{.}\) Let us write
\begin{gather*}
f(T,\theta_i\omega^n)=f(0,\theta_i\omega^n)+\sum_{j=1}^\infty a_j(i)T^j
\end{gather*}
where \(a_j(i)\in\Z_p\text{.}\) Since \(\chi_1\equiv\chi_2\pmod{\mf{p}}\) and \((1+p)\) is a unit in \(\mathcal{O}_m\text{,}\) we get that
\begin{gather*}
(\zeta_{\chi_1}(1+p)^{1-n}-1)^j\equiv(\zeta_{\chi_2}(1+p)^{1-n}-1)^j\pmod{\mf{p}}
\end{gather*}
for all \(j\geq1\text{.}\) Therefore,
\begin{gather*}
f(\zeta_{\chi_1}(1+p)^{1-n}-1,\theta_1\omega^n)\equiv f(\zeta_{\chi_1}(1+p)^{1-n}-1,\theta_2\omega^n)\pmod{\mf{p}}.
\end{gather*}
By definition of \(\chi_{i,n}\text{,}\) \(\zeta_{\chi_{i,n}}=\chi_{i,n}(1+p)^{-1}=\chi_i(1+p)^{-1}=\zeta_{\chi_i}\text{.}\) Therefore,
\begin{align}
f(\zeta_{\chi_i}(1+p)^{1-n}-1,\theta_i\omega^n)\amp =L_p(1-n,\chi_{i,n})\notag\\
\amp =(1-\chi_i\omega^n\omega^{-n}(p)p^{n-1})L(1-n,\chi_i)\notag\\
\amp =(1-\chi_i(p)p^{n-1})L(1-n,\chi_i).\tag{4.17}
\end{align}
Therefore, for \(n=1\text{,}\) we have
\begin{gather}
(1-\chi_1(p))L(0,\chi_1)\equiv(1-\chi_2(p))L(0,\chi_2)\pmod{\mf{p}}\tag{4.18}
\end{gather}
and for \(n\gt 1\text{,}\)
\begin{align*}
L(1-n,\chi_1)\amp \equiv (1-\chi_1(p)p^{n-1})L(1-n,\chi_1)\\
\amp \equiv (1-\chi_2(p)p^{n-1})L(1-n,\chi_2)\\
\amp \equiv L(1-n,\chi_2)\pmod{\mf{p}}.
\end{align*}
This completes the proof.
