The summation of elements of order $r$ at the finite field Consider the finite filed $GF(p^q)$ such that has elements of order $r$ that means $r\mid p^r-1$. Suppose that $a>1$ is a element of $GF(p^q)$ in which it's order is $r$ that means $r$ is the smallest number that holds at the relation $a^r=1$. We define elements $a_i$, $1\leq i \leq r-1$ in the following form:
$$
a_i=a^i \quad , \quad 1\leq i \leq r-1 \, .
$$
It can be proved that elements $a_i$, $1\leq i \leq r-1$ are distinct. Now, my question is that how to compute $\sum_{i=1}^{r-1}\,a_i$?
My try: 
$$
\begin{array}{l}
a_1+a_2+\cdots+a_{r-1}=a+a^2+\cdots+a^{r-1}\\
\\
=\underbrace{1+a+a^2+\cdots+a^{r-1}}+p-1=\frac{a^r+p-1}{a+p-1}+p-1\\
\\
=\frac{1+p-1}{a+p-1}+p-1==\frac{0}{a+p-1}+p-1=p-1
\end{array}
$$
Is my answer correct and then Is there another method to compute $\sum_{i=1}^{r-1}\,a_i$?
Thanks for any suggestions. 
 A: Here's how I address this problem:
first, I think it's worth demonstrating that the $a^j$, $1 \le j \le r - 1$, are distinct.  Suppose we had $1 \le j, k \le r -1$, $j \ne k$, and
$a^j = a^k; \tag{1}$
then we can assume without loss of generality that $j < k$.  Thus
$a^{k - j} = 1 \tag{2}$
with $0 < k - j < r - 1$.  But this contradicts the assumption that $r$ is the smallest such integer; so the $a^j$ are distinct for $1 \le j \le r - 1$.
Second, I think our OP Amin235's supposition that "$a > 1$" isn't really feasible in the context of finite fields, so I will suppose $a \ne 1$, which is meaningful, and a strong enough hypothesis to allow the calculation to proceed.
Having said these things, we proceed as follows:
set
$\sigma = \sum_1^{r - 1} a^j; \tag{3}$
then
$1 + \sigma = 1 + \sum_1^{r - 1} a^j = \sum_0^{r - 1} a^j, \tag{4}$
from which
$(a - 1)(1 + \sigma) = (a - 1) \sum_0^{r - 1} a^j = \sum_1^r a^j -  \sum_0^{r - 1} a^j = a^r - 1 = 0, \tag{5}$
since $a^r = 1$ by hypothesis.  Now since $a \ne 1$, $a - 1 \ne 0$, whence
$1 + \sigma = 0, \tag{6}$
or
$\sum_1^{r - 1} a^j = \sigma = -1, \tag{7}$
and the requisite sum is obtained.
