_Posted on June 04, 2025_ > [!theorem] Let $[a]_p \in \mathbb{Z}/p\mathbb{Z}$ , where $p \in \mathbb{Z}$ is prime, and assume that $p \nmid a$. Then $[a^{p-1}]_p = [1]_p$ . Equivalently, $[a^p]_p = [a]_p$. ^4baa36 Proof: Let $[a]_p \in \mathbb{Z}/p\mathbb{Z}$ and $p \in \mathbb{Z}$ be prime, where $p \nmid a$. We denote $S = \{[a]_p,[2a]_p,\cdots,[(p-1)a]_p\}$the set of non-zero multiples of $[a]_p$. Since $p$ is prime, $\mathbb{Z}/p\mathbb{Z}$ is a field and so cancellation holds in $\mathbb{Z} /p\mathbb{Z}$. Therefore, all the elements of $S$ are distinct, and since no element of $S$ is $[0]_p$, we have $S = \mathbb{Z}/p\mathbb{Z} \setminus \{[0]_p\}$ . Therefore, $[a]_p\cdot [2a]_p \cdots [(p-1)a]_p = [1]_p\cdot[2]_p\cdots[p-1]_p.$That is, $[(p-1)!\cdot a^{p-1}]_p=[(p-1)!]_p,$which implies (by cancellation), that $[a^{p-1}]_p=[1]_p.$Now, we want to show that this is equivalent to the statement that $[a^p]_p = [a]_p$. For the forward implication, simply multiply both sides by $[a]_p$. For the other direction, suppose that $[a^p]_p = [a]_p$. This implies that $[a^{p-1}\cdot a]_p = [1 \cdot a]_p.$ By cancellation, $[a^{p-1}]_p = [1]_p$. $\blacksquare$