$p$ prime, $p\mid a^k \Rightarrow p^k\mid a^k$ Suppose $p$ is a prime and $a$ and $k$ are positive integers. Prove that, if $p \mid a^k$, then $p^k\mid a^k$ also.
I have already proven that if $a,b,n\in\mathbb{N}$ and if $a^n\mid b^n$, then $a\mid b$. I tried using induction:
$k=1 \Rightarrow p\mid a^1 \Rightarrow p^1\mid a^1$. Assume the statement holds for some $k$. Then if $p\mid a^{k+1}\Rightarrow p\mid a^ka$. Since $p$ is prime, $p\mid a$ or $p\mid a^k$. If $p\mid a$, we are done. If $p\mid a^k$ then by assumption, $p^k\mid a^k$ and so by the lemma, $p\mid a$ and so $p^{k+1}\mid a^{k+1}$.
Does this work?

Hint: Try to prove that: $p\:|\:a^k$ implies $p\:|\:a$.
