TITLE: Why is the order of a $k$-cycle $\sigma$ equal to $k$?
QUESTION [8 upvotes]: 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.

REPLY [6 votes]: Denote $\sigma=(a_0,a_1,\ldots,a_{k-1})$ (to simplify notation later on).
Proposition: for all $n\in\Bbb N$ and $j\in\Bbb Z/k\Bbb Z$ it holds that $\sigma^n(a_j)=a_{j+n\pmod{k}}$.
Proof: by induction, since trivally $\sigma(a_{j+n\pmod{k}})=a_{j+n+1\pmod{k}}$.
Now note that by our proposition $\sigma^k=\mathrm{Id}$, and for all $j<k$ we have $\sigma^j\ne\mathrm{Id}$.