# 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.