I have it that $x^7 = x$ for all $a$ in a field $\mathbb{F}$. I want to prove that the characteristic of $\mathbb{F}$ must be 7.

Since $x^7=x$ implies $x^7 \equiv x (\mod p )$ I was trying to use Fermat's little theorem but I think it's only superficially similar. For it to work logically you would need to prove some kind of converse to Fermat's little theorem which I don't think would work out.

I did find a theorem online:

Let $G$ be a finite abelian multiplicative group. For all $a \in G$ the order of $a$ divides the order of $G$. In particular $a^{|G|} = 1$.

As a corollary they get:

The order of any element of a finite field with $q$ elements divides $(q-1)$.

Any element $x$ in a finite field of order $q$ satisfies $x^q = x$.

So I'm pretty sure I understand the theorem. We know that the set $\{a^i\}_{i=1}^q$ is a cyclic subgroup of the multiplicative group and we can apply Lagrange's theorem to deduce that $|\langle a \rangle | = q |$ divides $|G|$.

I don't understand the first corollary and that's probably why I don't understand the second corollary. So I can't solve my problem. Can someone explain to me why $q$ divides $(q-1)$?