If $\sigma = (a_1,a_2,\ldots,a_k)$, then the order of $\sigma$ is $k$ because $\sigma^k = Id$. I've tried finding a proof on the internet, but all sources just say "it's clear that", etc. I'm probably missing something really obvious. Can anyone tell me why this is?
Edit: I understand the intuition --- what I'm struggling with is a proof, a formal understanding of why this is true.