A permutation in $S_n$ is 'regular' if and only if it is a power of an n-cycle? 
Define a permutation $\alpha\in S_n $ to be regular if either $\alpha$ has no fixed points and it is the product of disjoint cycles of the same length, or $\alpha=(1)$.
Prove that $\alpha$ is regular if and only if $\alpha$ is a power of an n-cycle.

It is a homework question in J.J.Rotman's book: A first course in abstract algebra with applications.
I have just begun reading the book, and I find this question confusing----I've tried using induction, but couldn't figure out a good way. I would really appreciate your help.
 A: First, note that a power of an $n$-cycle is necessarily regular: if $\sigma=(a_0,a_2,\ldots,a_{n-1})$, then $\sigma^k$ maps $a_i$ to $a_{i+k\bmod n}$. If $\gcd(k,n)=1$, then $\sigma^k$ is an $n$-cycle again, hence regular. If $\gcd(k,n)=d$, then you should verify that $\sigma^k$ is a product of $d$ cycles, each of length $\frac{n}{d}$. This can be done by thinking about the additive cyclic group of order $n$ and the cosets of the subgroup generated by $d$.
For the converse, suppose that $\alpha = (a_0,\ldots,a_{d-1})(a_{d},\ldots,a_{2d-1})\cdots(a_{(k-1)d},\ldots,a_{kd-1})$ where $kd=n$. with the experience of the "if" part, you should be able to cook up an $n$-cycle $\sigma$ such that $\sigma^d = \alpha$.
To give you an example. Let's take $n=6$. Let $\sigma=(a_0,a_1,a_2,a_3,a_4,a_5)$. The powers of $\sigma$ are:
$$\begin{align*}
\sigma &= (a_0,a_1,a_2,a_3,a_4,a_5) &(\gcd(1,6)=1)\\
\sigma^2&= (a_0,a_2,a_4)(a_1,a_3,a_5) &(\gcd(2,6)=2)\\
\sigma^3&= (a_0,a_3)(a_1,a_4)(a_2,a_5) &(\gcd(3,6)=3)\\
\sigma^4 &= (a_0,a_4,a_2)(a_1,a_5,a_3) &(\gcd(4,6)=2)\\
\sigma^5&= (a_0,a_5,a_4,a_3,a_2,a_1) &(\gcd(5,6)=1)\\
\sigma^6 &= \mathrm{id}.
\end{align*}$$
and you can see that each power is regular.
Now, conversely, suppose we are given, say, $\alpha=(1,4)(2,3)(5,6)$. If it is a power of a $6$ cycle, it must be a cube of a $6$-cycle. Which $6$-cycle? There are a couple of possibilities; one is $(1,2,5,4,3,6)$.
