The additive topological group of Zp is topologically generated by 1, although it is not abstractly cyclic

The subgroup generated by $1$ consists exactly of the integer multiples $n\cdot 1$, so it is the canonical image of $\mathbb{Z}$ in ${\mathbb{Z}}_p$. Therefore [L1] says that its closure is all of ${\mathbb{Z}}_p$.

For every binary sequence $ϵ=({ϵ}_i{)}_{i\ge 0}\in \{0,1{\}}^{\mathbb{N}}$, define a compatible tuple $x(ϵ)$ by $x(ϵ{)}_n:={\sum }_{i=0}^{n-1}{ϵ}_i{p}^i(\mathrm{m}\mathrm{o}\mathrm{d}{p}^n)\mathrm{.}$ If two binary sequences first differ at index $j$, their $(j+1)$-st coordinates differ because their difference is divisible by $p^j$ but not by $p^{j+1}$. Hence $ϵ\mapsto x(ϵ)$ is injective.

Cantor's diagonal argument shows that $\{0,1{\}}^{\mathbb{N}}$ is uncountable, so step 1.2 makes ${\mathbb{Z}}_p$ uncountable. Every abstract cyclic group is the image of the countable group $\mathbb{Z}$ under $n\mapsto n\gamma$ for one generator $\gamma$, and is therefore countable. Hence the additive group of ${\mathbb{Z}}_p$ cannot be cyclic.

Step 1.1 proves topological generation by $1$, while steps 1.2 and 2.1 show that no element can generate the additive group abstractly.