Roots of an equation over the finite field $\operatorname{GF}(p^q)$ Consider the following  equation over the finite field $\operatorname{GF}(p^q)$ 
such that $r \mid p^q-1$:
\begin{align}
x^r=y^r \tag{1} \\
\end{align}
The solutions of $(1)$ over $\operatorname{GF}(p^q)$ are:
\begin{align}
x=\gamma^i\, y \quad , \quad 0\leq i \leq r-1\\ \tag{2}
\end{align}
where $\gamma$ is the element of order $r$ over $\operatorname{GF}(p^q)$.
It can be proved that $\gamma_i$'s are distinct elements. 
Another method to obtain solutions of $(1)$ is that considering the following equation:  
$$
x^r=y^r \Rightarrow
(x-y)\,g(x,y)=0 \tag{3}
$$
then find roots of the $g(x,y)$ over  $\operatorname{GF}(p^q)$,
where $g(x,y)$ is a polynomial of degree $r-1$ and based on the variables $x$ and $y$.
My question: How to prove that $g(x,y)=\prod_{i=1}^{r-1} (x-\gamma^i\, y)$? 
My try: We know by Newton's identities the following relation
$$
\prod_{i=1}^{r-1} (x-\gamma^i\, y)=\sum_{k=0}^{r-1}(-1)^{r-1-k}
e_{r-1-k}\,y^{r-1-k}\,x^k\tag{4}
$$
In addition, it can be proved that for $1\leq k\leq r-1$, we have
$$
p_k(\gamma^1,\cdots,\gamma^{r-1})=\sum_{i=1}^{r-1}\gamma^{ik}=-1\tag{5}
$$
There is a relation between $p_k$ and $e_k$ by Newton's identities that results that 
$$
e_{2k}(\gamma^1,\cdots,\gamma^{r-1})=1 \quad, \quad e_{2k-1}(\gamma^1,\cdots,\gamma^{r-1})=-1 \quad, \quad 1\leq k \leq \frac{r-1}{2}\tag{6}
$$
Therefore, by using $(5)$ and $(6)$ in the relation $(4)$ we conclude that 
$$
\prod_{i=1}^{r-1} (x-\gamma^i\, y)=\sum_{k=0}^{r-1}y^{r-1-k}\,x^k=g(x,y)
$$
Is it a correct proof? 
Thanks for any suggestions. 

I would just use the factorization
$$f(x)=x^r-1=\prod_{i=0}^{r-1}(x-\gamma^i).$$
It implies that
$$
x^r-y^r=y^rf(x/y)=y^r\prod_{i=0}^{r-1}(x/y-\gamma^i)=\prod_{i=0}^{r-1}(x-y\gamma^i).
$$
You can then cancel the factor $x-y$ corresponding to $i=0$.
The trick is known as homogenization. It adds one more variable to a polynomial, and gives, as an end product, a homogeneous polynomial, i.e. a polynomial such that all the terms share the same total degree.
