Let $\mathbb Z_p$ denote the set of $p$-adic intergers and $\mathbb Q_p$ the set of $p$-adic numbers. Write $\mu_{p^\infty}:=\{\zeta\in\overline{\mathbb Q}_p\mid \zeta^{p^n}=1\text{ for some }n\geqslant 0\}$. For $\Phi(T)\in\mathbb Z_p[[T]]$, I am stuck in proving the following
Problem. If there is some infinite subset $\Sigma\subseteq\mu_{p^\infty}$ such that $\Phi(\zeta-1)\in\mu_{p^\infty}$ for all $\zeta\in\Sigma$, then $\Phi(T)=\zeta_0(T+1)^b$ for some $\zeta_0\in\mu_{p^\infty}$ and $b\in\mathbb Z_p$.
Let's fix a $\zeta'\in\Sigma$. If we define a new power series $\Psi'(T)=\Phi(\zeta'(T+1)-1)$, then $\Psi'(0)=\Phi(\zeta'-1)\in\mu_{p^\infty}$. We define $\zeta_0':=\Phi(\zeta'-1)$, $\zeta_0:=\zeta_0'\zeta'^{-1}$ and $\Psi(T):=\Psi'(T)/\zeta_0\in\mathbb Z_p[[T]]$. The power series $\Psi(T)$ satisfies $\Psi(0)=1$ and if $\Psi(T)=(T+1)^b$, then $\Psi'(T)=\zeta_0'(T+1)^b$ and thus $\Phi(T)=\zeta_0(T+1)^b$. Hence the case is reduced as in the title, i.e. we may assume that $\Phi(0)=1$ and it remains to show that $\Phi(T)=(T+1)^b$ for some $b\in\mathbb Z_p$. This is where I am stuck, as I cannot see how it can be proved that the coefficients of $\Phi$ are exactly the binomial coefficients.
Any hint is appreciated.
For $p$ odd then $\zeta_0=1$. We are told that $\Phi(\zeta_{p^{n_j}}-1)=\zeta_{p^{m_j}}^{a_j}$ for a strictly increasing sequence $n_j$. Since $|\phi(\zeta_{p^{n_j}}-1)-1|\le |\zeta_{p^{n_j}}-1|$ we get $m_j\ge n_j$ and $\Phi(\zeta_{p^{n_j}}-1)=\zeta_{p^{n_j}}^{b_j}$.
Since $\overline{\Bbb{Z}_p}/(p)$ is a commutative ring of characteristic $p$ we have $(u+v)^p=u^p+v^p\bmod p$ which means that
$$\Phi(\zeta_{p^{n_j}}^{p^k}-1)\equiv \Phi(\zeta_{p^{n_j}}-1)^{p^k}\bmod p$$ Therefore $\zeta_{p^{n_j}}^{b_j}\equiv \zeta_{p^{n_{j+1}}}^{b_{j+1} p^{n_{j+1}-n_j}}\bmod p$ which implies that $ b_{j+1}\equiv b_j\bmod p^{n_j}$. Whence $$\Phi(\zeta_{p^{n_j}}-1)= \zeta_{p^{n_j}}^b $$ where $b=\lim_{j\to \infty} b_j\in \Bbb{Z}_p$.
I don't know if it is what you want to prove or if you expect something stronger.