Proof. If $n=1$, from $p \nmid a_{1}$ we know that when $y$ runs through a complete residue system modulo $p$, $a_{1} y$ also runs through a complete residue system modulo $p$ (see "Elementary Number Theory" $\Pi, p$. 6, Lemma 7). Therefore, there must be a natural number $b$ such that
$$a_{1} b \equiv 1(\bmod p)$$

It is easy to verify directly that in this case, (1) has exactly one solution $(\bmod p)$, and this solution is $x \equiv -b a_{0}(\bmod p)$.

Now assume $n>2$, and that for any polynomial of degree $<n-1$ with leading coefficient not divisible by $p$, the conclusion to be proved holds. We consider any $n$-degree polynomial $f(x)$ that meets the conditions.

Case one. If (1) has no solutions, then the conclusion of the theorem already holds for $f(x)$.
Case two. If (1) has a solution $x \equiv a(\bmod p)$, by polynomial division, there must be a constant $r_{1}$ and an $(n-1)$-degree integer polynomial $f_{1}(x)$ such that
$$f(x)=(x-a) f_{1}(x)+r_{1}$$

From $f(a) \equiv 0(\bmod p)$, we immediately get $r_{1} \equiv 0(\bmod p)$, i.e.,
$$f(x) \equiv (x-a) f_{1}(x)(\bmod p)$$

From (2), it is easy to see that the leading coefficient of $f_{1}(x)$ is still $a_{n}$, and $p \nmid a_{n}$. By the induction hypothesis, the congruence equation
$$f_{1}(x) \equiv 0(\bmod p)$$

has at most $n-1$ solutions $(\bmod p)$ (counting multiplicities), so the conclusion of the theorem also holds for any $n$-degree integer polynomial $f(x)$ that meets the conditions. This completes the proof of the theorem.