Question on Iwasawa Theory

Somewhat more concretely, you can show $T$ is topologically nilpotent as follows. We need to show that, given any $a\in A$, one has $T^na=0$ for $n\gg 0$. Fix some $a\in A$ and note that, since $\Gamma$ acts contintuously and $A$ is discrete, we have $A=\cup A^{\Gamma_n}$ where $\Gamma_n=\Gamma^{p^n}$ is a system of open neighborhoods of the identity. So there is some $n_0$ such that $a\in A^{\Gamma_{n_0}}$. In fact, $a\in A^{\Gamma_{n}}$ for all $n\geq n_0$. (This is because $\Gamma_n\supseteq \Gamma_{n+1}$, so $A^{\Gamma_n}\subseteq A^{\Gamma_{n+1}}$.) Thus, $a\in A^{\Gamma_n}$ for $n\gg 0$, so if $\gamma$ is a topological generator of $\Gamma$, we have $\gamma^{p^n}a=a$ for $n\gg0$. Since $\gamma$ acts by $1+T$ this gives $$ \big((1+T)^{p^n}-1)a=0,\quad n\gg0. $$ But $((1+T)^{p^n}-1)$ is a distinguished polynomial (i.e., the degree $p^n$ coefficient is 1 and all other coefficients are divisible by $p$). So $\big((1+T)^{p^n}-1\big)=pF(T)+T^{p^n}$ for some polynomial $F(T)\in \mathbb{Z}_p[T]$. Hence \begin{equation} F(T)(pa)+T^{p^n}a=0,\quad n\gg0. \end{equation} We can now induct on the order of $a$. If $a$ has order $p$ then we're done by the above equality. Suppose the result holds for elements of order $p^{N-1}$ and let $a$ have order $p^N$. Then $pa$ has order $p^{N-1}$ so $T^n(pa)=0$ for $n\gg0$. Applying $T^n$ to the above equality finishes the proof.