If $p \mid a^n$ then does $p^n \mid a^n$? I'm trying to figure out if the statement is true or not and I need to prove it if so.
Let $p$ be a  prime and $a$ be an integer.
If $p\mid a^n$ , is it true that $p^n\mid a^n$ ?
I'm not sure how i would approach this problem.
I started as $a^n = p \cdot b$ for some $b \in \mathbb{Z}$.
 A: This can be done using a basic property.
Note that if $p|xy$ then $p|x$ or $p|y$.
So $p|a\cdot a^{n-1}$ means $p|a$ (when we are pretty much done) or $p|a^{n-1}$.
So there is a simple induction to show $p|a$ and we are done as before
A: Since $a$ is an integer we may write the canonical prime factorization of $a$ as $a=p_1^{k_1}\cdots p_s^{k_s}$. Then $a^n=(p_1^{k_1}\cdots p_s^{k_s})^n=p_1^{nk_1}\cdots p_s^{nk_s}$. So if $p\in \{p_1,...,p_s\}$, in other words $p|a$, then $p^n|a^n$.
Another proof: Suppose $p|a^n$. Then $p|a(a^{n-1})$. So $p|a$ or $p|a^{n-1}$. Continue by induction to find $p|a$. Thus $a=pb$ for some integer $b$, and $a^n=p^nb^n$. Therefore, $p^n|a^n$.
A: This is easier if you put in an intermediate statement: $p\mid a^n\implies p\mid a\implies p^n\mid a^n$ (this is used in the other answers as well, I just wanted to highlight this a bit more). The first implication follows by induction from Euclid's lemma, the second from the general fact that $(a\mid k)\land (b\mid l)\implies ab\mid kl$.
