We will show $$a^{p} \equiv a \pmod{p}$$ with $p$ prime. The equivalent statement $$a^{p-1}\equiv 1 \pmod{p}$$ when $p$ does not divide $a$ follows by cancelling $a$ both sides (which can be done since then $a,p$ are coprime).

When $a=1$ , we have $$ 1^{p} \equiv 1 \pmod{p}$$

Now assume the theorem holds for some positive $a$ and we want to prove the statement for $a+1$ . We will have as a direct consequence that $$a^p \equiv a \pmod{p}$$

Let's examine $a+1$ . By the binomial theorem, we have \begin{eqnarray*} (a+1)^{p} & \equiv & \binomial{p}{0}a^p + \binomial{p}{1}a^{p-1} + \cdots + \binomial{p}{p-1} a + 1 \\ & \equiv & a + pa^{p-1} + p\frac{(p-1)}{2}a^{p-2} + \cdots + p a + 1 \\ & \equiv & (a + 1) + [ pa^{p-1} + p \frac{(p-1)}{2}a^{p-2} + \cdots + p a ] \end{eqnarray*} However, note that the entire bracketed term is divisible by $p$ , since each element of it is divsible by $p$ . Hence $$(a+1)^p \equiv (a+1) \pmod{p}$$

Therefore by induction it follows that $$ a^p\equiv a \pmod{p $$ for all positive integers $a$ .

It is easy to show that it also holds for $-a$ whenever it holds for $a$ , so the statement works for all integers $a$ .