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 $(1,2,5,4,3,6)$.