Proof

By hypothesis, $p$ is a prime number and $f(x)=a_0+a_1x+\ldots+a_nx^n$ a polynomial of the degree $n$ with integer coefficients $a_0,\ldots,a_n$ and with $p\not\mid a_n$.
If $n=0$ then since $p\not\div a_0$, we have $a_0(p)\not\equiv 0(p)$ and the polynomial has no roots.
Let $n > 0$ and let the claim be correct for $n-1.$
Assume, $f(x)(p)\equiv 0(p)$ had more than $n$ roots.
- In particular, assume $x_0,x_1,\ldots,x_n$ are $n+1$ roots. Then $$\begin{array}{rcl} f(x)-f(x_0)&=&\sum_{r=1}^na_i(x^r-x_0^r)\\ &=&(x-x_0)\sum_{r=1}^na_r(x^{r-1}+x_0x^{r-2}+\ldots+x_0^{r-2}x+x_0^{r-1})\\ &=:&(x-x_0)g(x) \end{array}$$ with some polynomial $g(x)=b_0+b_1x_1+\ldots+b_{n-1}x^{n-1},$ such that $b_{n-1}=c_n$ and $p\not\mid b_{n-1}.$
- Therefore, for $k=1,\ldots,n$ we have $$(x_k-x_0)g(x_k)\equiv f(x_k)-f(x_0)\equiv 0-0\equiv 0\mod p.$$
- It follows that $g(x_k)(p)\equiv 0(p)$ would have $n$ roots for $k=1,\ldots,n$, in contradiction to the result that it has only at most $n-1$ roots.
Therefore, the assumption $f(x)(p)\equiv 0(p)$ had more than $n$ roots is wrong.
Thus, $f(x)(p)\equiv 0(p)$ has at most $n$ roots.
∎

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