group theory - Explanation to Fermat's little theorem proof

Fermat's little theorem
$\forall a \in \mathbb{Z}$ and every prime p.
Then, $a^{p}\equiv a\pmod p$

$a=pm+r$

$\forall 0 \leq r

Proof for $r\not\equiv 0:$

Then, $\forall r \in \bar{U}\left ( p \right )$ and $\left | \bar{U}\left ( p \right ) \right |=p-1$

$r^{\left | \bar{U}\left ( p \right ) \right |}=e$ by a certain theorem in Cosets.

But this is really just $r^{p-1} \equiv 1\pmod p$

How does the last equivalence follows?
My knowledge of number theory is almost non-existant.
A verbose explanation would really help.

Thanks in advance.