suppose that |σ| = k, and that σ = (a1 a2....am)
then σn sends aj to aj+n (mod m)
since σk = id, we have:
aj = aj+k (mod m), for all j. in particular, 1 = 1+k (mod m) which means that k is a multiple of m.
since k is the least positive power of σ that equals the identity, 1+k is the smallest number > 1 such that 1 = 1+k (mod m), which shows that k = m, so σ is a k-cycle.
if, on the other hand, σ is a k-cycle, then for n < k:
σn sends a1 to a1+n (mod k), and since n < k, n+1 ≤ k, so σn(a1) = an+1 ≠ a1, so σn is not the identity.
however, σk(aj) = aj+k (mod k) = aj, so σk = id, thus |σ| = k.
(we need k ≥ 2 so that σ1 is not a 1-cycle, since 1-cycles ARE the identity).