First, $0^p\equiv 0\pmod{p}$ and $1^p\equiv 1\pmod{p}$ plainly hold.

By the conclusion of # 4233, we have $$(x+y)^p\equiv x^p + y^p\pmod{p}$$

Setting $x=y=1$ leads to $$2^p\equiv 1^p + 1^p\equiv 2\pmod{p}$$

Setting $x=1$ and $y=2$ leads to $$3^p\equiv 1^p + 2^p\equiv 1 + 2\equiv 3 \pmod{p}$$

Repeating this process will lead to the conclusion that $$a^p \equiv a\pmod{p}$$

holds for all $a=0, 1 2, 3, \cdots, p-1$. For any $a \ge p$, $a\pmod{p}$ msut equal to one of $\{0, 1, \cdots, p-1\}$. Hence we conclude $a^p\equiv a\pmod{p}$ holds for any $a$.