Let $a\in Z_p^{*}$. I need to prove the sequence $\{a^{p^n}\}_{n\geq 0}$ converges in $\mathbb{Z}_p$ to $w$, where $$a=wb$$ for $w$ is the $p-1$ root of unity in $\mathbb{Z}_p^*$ and $b\in 1+p\mathbb{Z}_p$.
I feel like I need to use Hensel's Lemma for this question but I do not know how to start it. Is there any particular hint for this question?
On
$\Bbb{Z/p^k Z}^\times$ has $p^{k-1}(p-1)$ elements. Thus $order(a\bmod p^k)$ divides $p^{k-1}(p-1)$,
For $n\ge k$, $order(a^{(p-1)p^n}\bmod p^k)= 1$ ie. $$a^{p^{n+1}}\equiv a^{p^n}\bmod p^k$$
This implies that the limit converges $$w=\lim_{n\to \infty} a^{p^n}\in \Bbb{Z}_p$$
Also $a\equiv w\bmod p$ and $w^p = w$ so that $w$ is the unique $p-1$-th root of unity such that $w^{-1}a\in 1+p\Bbb{Z}_p$.
On
We already know from Hensel's lemma that there's an element $w$ that satisfies $w^p=w$. Further we know that $a$ shares at least one digit in common with it, so we can write for some $k\ge 1$ and $|z_k|_p \le 1$,
$$a = w + p^k z_k$$
Raising this to the $p$th power and expanding the binomial,
$$a^p = w^p +\sum_{k=1}^p \binom{p}{n} (pz_k)^n w^{p-n}$$
Simplifying $w^p=w$ and noticing we have at least a factor of $p^{n+1}$ we can factor out, for some other p-adic integer with $|z_{k+1}|_p \le 1$
$$a^p = w + p^{k+1} z_{k+1}$$
In essence raising it to the $p$th power has "uncovered" a digit of $w$. So we can repeat this process make the sequence of terms now,
$$a^{p^n} = w + p^{k+n} z_{k+n}$$
$$\lim_{n \to \infty} a^{p^n} = \lim_{n \to \infty} w + \lim_{n \to \infty} p^{k+n} z_{k+n} = w$$
Because the $z_k$ are bounded and $p^k\to 0$ and the actual terms of the $z_k$ sequence are irrelevant.
Don't use Hensel's lemma: use the contraction mapping theorem. For $a, b \in \mathbf Z$, show $$ a \equiv b \bmod p^k \Longrightarrow a^p \equiv b^p \bmod p^{k+1}. $$ Since $p$-adic integers modulo $p^k$ are "just like" integers modulo $p^k$, the same implication holds for $a, b \in \mathbf Z_p$. Therefore when $a$ and $b$ are in $\mathbf Z_p$, $$ |a^p - b^p|_p \leq \frac{1}{p}|a-b|_p $$ Indeed, if $a = b$ there is nothing to check, and if $a \not= b$ let $p^k$ be the power of $p$ where $a - b = p^ku$ for $u \in \mathbf Z_p^\times$. That makes $|a - b|_p = 1/p^k$, so from the above implication, $$ |a^p - b^p|_p \leq \frac{1}{p^{k+1}} = \frac{1}{p}|a-b|_p. $$
Now use the contraction mapping theorem for the $p$th power map on the coset $C = a + p\mathbf Z_p$. This is a complete metric space and the $p$th power map sends $C$ to itself since $x^p \equiv x \bmod p\mathbf Z_p$ for all $x \in \mathbf Z_p$. The limit coming from this contraction is a number $w \in C$ such that $w^p = w$. If $a \in \mathbf Z_p^\times$ then $w \not= 0$, so $w^{p-1} = 1$. (If $a \in p\mathbf Z_p$ then $w = 0$.)