I want to proof that $∑_{c∈F_{q}}f(c)^{q-1}=-1$

Question

[ Let $f\in \mathbb{F}_q[x]$, with $f$ has only one root over $\mathbb{F}_q$. I want to proof that $∑_{c∈F_{q}}f(c)^{q-1}=-1$][1]

[1]:

Answer 1

Hint:

without loss of generality let the single root of the polynomial be $c = 0$. (Otherwise shift the polynomial via $g(x) = f(x+r)$).

For $c \not= 0$, $f(c) \not= 0$, since this was your only root. Forget about the polynomial $f$ and just write $f(c) = g_c$, knowing that $g_c \in F^{\times}_q$.
Your sum becomes $$ \sum_{c\in F^{\times}_q} g_c^{q-1} $$ Now you should know a couple of things about $F^{\times}_q$. It's an Abelian group of order $q-1$. What can you say about the orders of the elements in such a group?