By arc
\(_p\)-descent and
Example 20.3.8, we may reduce to the case where
\(R\) is a product of
\(p\)-complete AIC valuation rings, and then to the case of a single such ring. In this case, the map
\(\phi^{-1}(d) A \stackrel{\phi/d - 1}{\to} A\) is surjective modulo
\(p\) (by the AIC property) and hence surjective by derived Nakayama (
Proposition 6.3.1).
Suppose that \(R\) is of characteristic \(p\text{.}\) In this case, we may take \(d = p\text{,}\) and then \(\ZZ_p(n)_{\lens} \cong \left( A \stackrel{\phi-p^n}{\to} A \right)\text{.}\) The map \(\phi-p^n\) on \(A\) is visibly injective, so both sides of the desired isomorphism are zero.
Suppose next that \(R\) is of characteristic \(0\text{.}\) Choose a morphism \(\ZZ_p[\mu_{p^\infty}] \to R\text{,}\) let \(\epsilon \in R^\flat\) be the element \((1,\zeta_p,\zeta_{p^2},\dots)\text{,}\) and put \(q = \epsilon^\sharp\text{.}\) We can then take the generator of \(d\) to be \([p]_q = (q^p-1)/(q-1)\text{;}\) we may then identify \(\ZZ_p(n)_{\lens}\) with \((q-1)^n \ZZ_p \subset \phi^{-1}(d^n) A\text{.}\) This gives the desired natural isomorphism \(\ZZ_p(n)_{\lens} \cong \ZZ_p(1)_{\lens}^{\otimes n}\text{.}\)
To specify a natural isomorphism \(\ZZ_p(n) \cong \ZZ_p(n)_{\lens}\text{,}\) it now suffices to do so for \(n=1\text{.}\) In this case, we must check that the action of \(\Gal(\QQ_p(\mu_{p^\infty}))\) on \((q-1)\ZZ_p\) matches the action on \(\lim_n \mu_{p^n}\text{;}\) this follows from the fact that
\begin{equation*}
q^m - 1 \equiv m(q-1) \pmod{d(q-1)} \qquad (m \in \ZZ).
\end{equation*}