Let \(f(x) = \sum_{i=0}^{m} a_i x^i\) and \(g(x) = \sum_{i=0}^{n} b_i x^i\text{.}\) Suppose that \(p\) is a prime dividing the coefficients of \(f(x) g(x)\text{.}\) Let \(r\) be the smallest integer such that \(p \notdivide a_r\) and \(s\) be the smallest integer such that \(p \notdivide b_s\text{.}\) The coefficient of \(x^{r+s}\) in \(f(x) g(x)\) is
\begin{equation*}
c_{r + s} = a_0 b_{r + s} + a_1 b_{r + s - 1} + \cdots + a_{r + s - 1} b_1 + a_{r + s} b_0.
\end{equation*}
Since \(p\) divides \(a_0, \ldots, a_{r-1}\) and \(b_0, \ldots, b_{s-1}\text{,}\) \(p\) divides every term of \(c_{r+s}\) except for the term \(a_r b_s\text{.}\) However, since \(p \mid c_{r+s}\text{,}\) either \(p\) divides \(a_r\) or \(p\) divides \(b_s\text{.}\) But this is impossible.