Nilpotent action on $p$-group Let $A$ be a finite, abelian $p$-group and $\Gamma$ is a multiplicative topological group isomorphic with the additive group of $p$−adic integers $\mathbb Z_p.$ and let $\gamma_0$ a topological generator of $\Gamma.$ Suppose that $\Gamma$ acts continuously  by multiplication by $\gamma_0.$ $$(a,\gamma_0)\longrightarrow \gamma_0.a.$$
How to prove that $T=1-\gamma_0$  acts nilpotently on $A $ (i.e:if $a\in A$ then $T^{n_0}a = 0$ for $n_0\in \mathbb N^*.$) ?
Thank you for any help.
 A: Let $a \in A$. $A$ carries the discrete topology, so the singleton set $\{a\}$ is open in $A$. 
Since the action map $\alpha : \Gamma \times A \to A$ is continuous, $\alpha^{-1}(\{a\})$ is open in $\Gamma \times A$. 
So $\alpha^{-1}(\{a\}) \cap ( \Gamma \times \{a\} ) = \{(\gamma,a) \in \Gamma \times A : \gamma \cdot a = a\}$ is open in $\Gamma \times \{a\}$.
Thus the stabiliser of $a$ is an open subgroup of $\Gamma$, and therefore has finite index in $\Gamma$. Since $\Gamma$ is pro-$p$, this index is a power of $p$; say $p^n$.
Thus $\gamma_0^{p^n} \cdot a = a$. So $( (1 + T)^{p^n} - 1 )\cdot a = 0$. Since $(1 + T)^{p^n} \equiv 1 + T^{p^n} \mod p$, we get $T^{p^n} \cdot a \in pA$. This is true for all $a \in A$ and $A$ is finite, so in fact $T^{p^m} \cdot A \subseteq pA$ for some $m$.
So $T^{2p^m} \cdot A \subseteq T^{p^m} \cdot pA = p T^{p^m} \cdot A \subseteq p^2 A$ and similarly $T^{ip^m} \cdot A \subseteq p^iA$ for all $i \geq 0$. But $A$ is a finite abelian $p$-group so $p^i A = 0$ for some $i$. Hence $T^{p^m i} \cdot A = 0$.
