If $a^k \equiv 1 (\text{mod} \ p) $, then obviously $p$ does not divide $a$. Therefore, by Fermat's little theorem we have:
$$ a^{p-1} \equiv 1 (\text{mod} \ p) \tag{A} $$
Say $k$ is the order of $a$ modulo $p$. That is, $k$ is the smallest positive integer such that $a^k \equiv 1 (\text{mod} \ p)$. So, we have
$$ a^k \equiv 1 (\text{mod} \ p) \tag{B} $$
And say, for the sake of contradiction, that $k$ does not divide $p-1$. So, we have
$$ p-1 = kq + r, \ \text{where} \ q \ \text{is an integer and } \ \ 1\le r \le k-1
\tag{C} $$
Using $(B)$ one can deduce that $ a^{kq} \equiv 1 (\text {mod} \ p) $. Using this and $(A)$ we get
$$ p \ | a^{p-1}- a^{kq} \implies p \ |a^{kq}(a^{p-1-kq} -1 ) \implies p|a^{kq}(a^{r}-1) $$
Since $p$ does not divide $a^{kq} \implies p|(a^{r}-1) \implies a^r \equiv 1 (\mod{p}) $. But $r<k$ and we had assumed $k$ to be the smallest positive integer for which $a^k \equiv 1 (\text{mod} \ p)$. So, we have reached a contradiction; therefore, our original assumption must be incorrect and $k$ divides $p-1$.
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