Let $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]] := \varprojlim_{n,m}(\mathbb{Z}/n)[x]/(x^m-1)$ be the complete group algebra of the profinite free group of rank 1. In Corollary 5.9.2 of Ribes-Zalesski's Profinite Groups, they state that $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]\cong \widehat{\mathbb{Z}}[[t]]$, citing a paper of Lim which doesn't seem to prove exactly what they claim (though I'm sure it must follow, if one is sufficiently familiar with the theory).

Here, I think we want $\widehat{\mathbb{Z}}[[t]]$ to be given the topology corresponding to the product topology on $\prod_{n\ge 0}\widehat{\mathbb{Z}}$, indexed by the coefficients of $t^n, n\ge 0$, though I could be mistaken.

Anyhow, I would like to show directly that $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]\cong \widehat{\mathbb{Z}}[[t]]$. If this is true, then the map should be given by identifying a generator "$x$" of the group algebra with $1+t$.

However, one difficulty is that I don't know of a neighborhood basis of open ideals of $\widehat{\mathbb{Z}}[[t]]$. At first I thought the ideals $(n,t^m)$ for $n,m\ge 1$ should suffice, and indeed for every such ideal, one can find some $N,M\ge 1$ such that the map "$x\mapsto 1+t$" defines a quotient map $$\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]\rightarrow(\mathbb{Z}/N)[x]/(x^M-1) \rightarrow \widehat{\mathbb{Z}}[t]/(n,t^m)$$ However when I try to go in the other direction, it seems what I need is - for every $N,M\ge 1$, to find an $n,m$ such that $t\mapsto x-1$ induces a map $$\widehat{\mathbb{Z}}[t]/(n,t^m)\rightarrow(\mathbb{Z}/N)[x]/(x^M-1) $$ However, this is clearly impossible, since $t$ is always nilpotent on the left, and yet $x-1$ is rarely nilpotent on the right.

Thus, I guess the ideals $\{(n,t^m)\}_{n,m\ge 1}$ do not form a neighborhood basis of 0 in $\widehat{\mathbb{Z}}[[t]]$, which leaves me stuck...