Prove an interesting result involving repetition when raising integers to a power, mod p. This problem was in my Elementary Number Theory book, but the problem didn't have a solution. Suppose we have a prime $p$. Let $a|p-1$. Then, the list $\{1^a,2^a,3^a,4^a,...,(p-1)^a\}$, contains $(p-1)/a$ distinct values, mod p. Furthermore, each value occurs $a$ times.
I'm not quite sure how to approach this. I'm thinking that there will be some FLT involved, but I don't know where I could implement this. My thinking was also that we could use something like $x^d|x^{p-1} \rightarrow x^{p-1} = k\cdot x^a$, and use this fact to count repetitions.
Any help would be appreciated.
 A: Since $p$ is a prime, there is an element $g$ which is a primitive root and, thus, also generator of the multiplicative group of non-zero remainders. This means each integer from $1$ to $p - 1$, inclusive, is congruent modulo $p$ to a unique power of $g$ from $1$ to $p - 1$. Thus, the list of $\{1^a,2^a,3^a,4^a,\ldots,(p-1)^a\}$, would be equivalent (i.e., congruent modulo $p$) to, in some order, $\{g^a, g^{2a}, g^{3a}, \ldots, g^{(k-1)a}\}$.
For any indices $i \lt j$, the $2$ elements being equivalent means
$$\begin{equation}\begin{aligned}
g^{ia} & \equiv g^{ja} \pmod{p} \\
1 & \equiv g^{(j - i)a} \pmod{p}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Since $g$ is a primitive root, it's only equivalent to $1$ when its power is an integral multiple of $p - 1$. This means there's an integer $k \gt 0$ where
$$\begin{equation}\begin{aligned}
(j - i)a & = k(p - 1) \\
j - i & = k\left(\frac{p-1}{a}\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
For $k = 1$, you get the first repetition of values, so there would be $\frac{p - 1}{a}$ distinct values in that range. As these set of values will then keep repeating among the $p - 1$ values in total, this means each value will repeat $\frac{p - 1}{\frac{p - 1}{a}} = a$ times.
