According to Euclidean lemma it is defined that if $p$ is prime then $$p|ab\Rightarrow p|a\lor p|b$$
How to prove by descending induction that if $$p|a^n \Rightarrow p|a $$
knowing that $a^n = a \times a^{n-1}$
If there is an easier solution don't hesitate to post
You could do it that way :
$$a^n=a^{n-1}*a\ ,$$
so either p divides a or $a^{n-1}$.
Continue until n=1.
Suppose that $p$ does not divide $a= p_1^{k_1} \cdots p_l^{k_l}$, this means that there is no prime divisor of $a$ that equals $p$. $a^n= (p_1^{k_1} \cdots p_l^{k_l})^n$, so $p$ is not a divisor of $a^n$ either. The theorem holds by contraposition.
