Prove by contradiction. If the conclusion does not hold, then there must be a $d, 0 \leqslant d \leqslant n$, such that $p \mid a_{j}$, $d < j \leqslant n$, and $p \nmid a_{d}$. In this case, the number of solutions to the congruence equation (2) is the same as the number of solutions to the congruence equation
$$a_{d} x^{d}+\cdots+a_{0} \equiv 0(\bmod p)$$

However, by Theorem 2, the number of its solutions $\leqslant d \leqslant n$. This is a contradiction. The proof of (ii) is left to the reader (the definitions of equivalence modulo $p$ and congruence modulo $p$ are given in Chapter 3, §1, Definition 2).