I have consulted some other sources, and think that I have some handle on the basic ideas behind the proof, but am having trouble articulating them.
I understand that $\sigma$ sends $a_k$ to $a_{k} + 1$ (mod m). Moreover, $\sigma (a_k +1) = a_k +2$ (mod m) That is to say $\sigma ^2 $ sends $a_k$ to $a_k + 2$. One can repeat this indefinitely and can easily show that for any $i \in \mathbb{Z}$$ \sigma^i (a_k) = a_k +i$ (mod m)
Is this the right direction or am I dead wrong?
It might help if you consider a simple example, such as the powers of the circular permutation $\sigma=(1,2,3,4,5,6)$. You'll see that $\sigma^2=(1,3,5)(2,4,6)$ consists of two disjoint 3-cycles, and that $\sigma^3$ consists of three disjoint 2-cycles.
Suppose $\sigma=(1,2,\ldots,m)$. Then $\sigma^i$ maps $k$ to $k+i$ for every $k$ (where sum is modulo $m$), hence maps $k+i$ to $k+2i$, $k+2i$ to $k+3i$, and so on. Thus $\sigma^i: k \mapsto k+i \mapsto k+2i \mapsto \cdots k+(m-1)i \mapsto k$. These elements are distinct (i.e. $\sigma^i$ is an $m$-cycle) iff $k+ix$ and $k+iy$ are distinct whenever $x$ and $y$ are distinct, i.e. iff $k+ix \not\equiv k+iy$, for all distinct $x,y \in \{0,1,\ldots,m-1\}$, i.e. iff $i(x-y) \not\equiv 0 \pmod m$. Show that this condition is satisfied iff $i$ and $m$ are relatively prime.