Let \(p\) be our prime. All integers, including \(a\), has a unique prime factorization:
$$\begin{gather} a = p_1^{e_1}p_2^{e_2} \ldots p_n^{e_n} \\ a^n = p_1^{ne_1}p_2^{ne_2} \ldots p_n^{ne_n} \end{gather}$$
If \(p|a^n\), then \(p\) has to be one of \(p_1,p_2,\ldots,p_n\) which means \(p|a\).