 Solution

A transposition in the symmetric group $S_n$ exchanges two distinct symbols and fixes all others. A permutation cycle $(a_1{a}_2\cdots a_k)$ sends $a_j$ to $a_{j+1}$, sends $a_k$ to $a_1$, and fixes the other symbols. Its entries are distinct. For $k=1$ this is the identity permutation.

A group element has order of a group element $k$ if $k$ is the least positive integer for which its $k$th power is the identity element. For the displayed permutation cycle, its $j$th power sends $a_i$ forward $j$ places, with subscripts read modulo $k$. Hence the $k$th power fixes every symbol, whereas for $0<j<k$ the image of $a_1$ is different from $a_1$. Thus its order of a group element is exactly $k$, including $k=1$.