Order of Cycle is Length of Cycle

Theorem
Let $S_n$ denote the symmetric group on $n$ letters.

Let $\pi \in S_n$ be a cyclic permutation of length k.

Then:
 * $\order \pi = k$

where:
 * $\order \pi$ denotes the order of $\pi$ in $S_n$.

Proof
Denote $\pi=(a_0,a_1,\ldots,a_{k-1})$. Observe that

$(j+n)\pmod{k}+1 = (j+n+1)\pmod{k}$

$\implies \pi(a_{(j+n)\pmod{k}})=a_{(j+n+1)\pmod{k}}$

$\implies \forall n\in\Bbb N$ and $\forall j\in\Bbb Z/k\Bbb Z$, $\pi^n(a_j)=a_{(j+n)\pmod{k}}$

$\implies \pi^k=\mathrm{Id}$ and $\pi^j\ne\mathrm{Id} \ \forall \ j < k$.