I have to show that for $\alpha \in \mathbb{Z}_p$, $\alpha^{p^n} \equiv \alpha^{p^{n-1}}\ \text{mod}p^n$ for all $n \in \mathbb{N}$.
I tried using mathematical induction, but I end up with something like $\alpha^{p^n} = (\alpha^{p^{n-1}})^p \equiv (\alpha^{p^{n-2}})^p \text{mod}p^{n-1} = \alpha^{p^{n-1}} \text{mod}p^{n-1} $
But clearly I'm missing something. Can I get some help? This is a part of Question 19 in the first Exercise of Koblitz's book on p-adic analysis
This can be proved using the following lemma: Suppose $\alpha, \beta \in \mathbb{Z}_p$, and $n \ge 1$. If $\alpha \equiv \beta \pmod{p^n}$, then $\alpha^p \equiv \beta^p \pmod{p^{n+1}}$. Proof: $\alpha^p - \beta^p = (\alpha-\beta) (\alpha^{p-1} + \alpha^{p-2} \beta + \cdots + \beta^{p-1})$. Now by hypothesis, $p^n \mid \alpha - \beta$; but also, since $n \ge 1$, we have $\alpha \equiv \beta \pmod{p}$, so then $\alpha^{p-1} + \alpha^{p-2} \beta + \cdots + \beta^{p-1} \equiv p \alpha^{p-1} \equiv 0 \pmod{p}$, so $p \mid \alpha^{p-1} + \alpha^{p-2} \beta + \cdots + \beta^{p-1}$.
Now, to apply this to an inductive proof of the desired statement: for $n=1$ it reduces to $\alpha^p \equiv \alpha \pmod{p}$ which is just Fermat's little theorem. For the inductive step, applying the lemma to the inductive hypothesis $\alpha^{p^n} \equiv \alpha^{p^{n-1}} \pmod{p^n}$ pretty much gives directly that $\alpha^{p^{n+1}} \equiv \alpha^{p^n} \pmod{p^{n+1}}$.