For prime $p$ and any $1 \le a \le (p-1)$ there exists $n > k \ge 1$ where $a^n \equiv a^k \pmod{p}$

Question

Please prove the following:

For a prime $p$ and any $1 \le a \le (p-1)$ there exists $n > k \ge 1$ where $a^n \equiv a^k \pmod{p}$

Answer 1

You can take $k=1$ and $n=p$ and use Fermat's little theorem.

Another proof (this works even $p$ is not a prime)(combinatorial): the sequence $(a^i)_{i\in \mathbb{N}}$ is infinite and the sequence of its remainders modulo $p$: $(a^i \mod p)_{i\in \mathbb{N}}$ can take only finite values $0,1,\cdots ,p-1$ as a consequence of pigeonhole principle there exists two integers $i<j$ such that $a^i \mod p= a^j \mod p $.