How can I look to the semidirect product $G_n$ of $A_n = \Bbb{F}_p[X]/(X^{p^n} - 1)$ by $\langle x_n \rangle \simeq \Bbb{Z}/p^n\Bbb{Z}$ with $x_n$ acting as multiplication by $X$?
I want to calculate the number of topological generators of $\varprojlim G_n$ w.r.t natural projections $A_{n+1} \to A_n$ and the rank, but is painfull work with the structure of the finite groups
UPDATE. I'm working with a slight modification by considering $G = \varprojlim C_p \wr C_{p^n}$. I think I can able to show that $G$ is topologically $2$-generated. The idea is to show that each $C_p \wr C_{p^n}$ is $2$-generated so use a result that ensures $\sup \{d(C_p \wr C_{p^n})\} < \infty$ implies $d(G) = d(C_p \wr C_{p^m})$ for $m$ sufficiently large. Now, there is a problem on Analytic Pro-$p$ groups asking to show that $G$ has infinite rank. I don't know how to show it.
UPDATE. Consider the subgroup $H_1 = C_p \rtimes 1$ of $C_p \wr C_{p^n}$. This is embedded in $C_p \wr C_{p^n}$ for every $n$. If we define $H_2 = (C_p \times C_p) \rtimes 1$, it is embedded in $C_p \wr C_{p^n}$ for every $n > 0$. Repeat the same processes to $H_m$. Each index set is cofinal in the original index of original inverse limit. Also, $H_m = \varprojlim_n H_m$ and $d(H_m) \to \infty$ as $m \to \infty$. This shows that rank of $G$ is infinite. I don't know if it is correct, but is was my better idea for now.
Each $G_n$ is generated by two elements, and hence so is (topologically) $G$.
Indeed, let $u_n=$"$1$" be the unit element of the ring $A_n$. Then the $A_n$ is generated as $G_n$-module by $u_n$. Indeed, $x_n^k$ conjugates $u_n$ to the ring element "$X^k$", and these generate $A_n$ additively. Hence, $\{x_n,u_n\}$ is a generating subset of $G_n$.