sequence $\{a^{p^{n}}\}$ converges in the p-adic numbers.

Let $a\in \mathbb{Z}$ be relatively prime to $p$ prime. Then show that the seqeunce $\{a^{p^{n}}\}$ converges in the $p$-adic numbers.

This to me seems very counter intuitive. Since $(a,p)=1$ the norm will always be $1$. I really have no idea what to do with $|a^{p^{n}}-a^{p^{m}}|$ factoring gets me nothing and we can’t use the nonarchimedian property because they have the same norm. Any hints or ideas would be great. Thanks.

Solutions Collecting From Web of "sequence $\{a^{p^{n}}\}$ converges in the p-adic numbers."

Recall that the Euler totient function has values $\phi(p^n)=p^{n-1}(p-1)=p^n-p^{n-1}$ for all $n$. This means that for all $a$ coprime to $p$ we have the congruence
$$
a^{p^n}\equiv a^{p^{n-1}}\pmod{p^n}.
$$
By raising that congruence to power $p^{m-n}$ a straightforward induction on $m$ proves that
$$
a^{p^m}\equiv a^{p^{n-1}}\pmod{p^n}
$$
for all $m\ge n$. This holds for all $n$ implying that the sequence is Cauchy, and thus convergent w.r.t. the $p$-adic metric.