I'm trying to prove that if a|b then $a^n|b^n, \forall n \in \mathbb{N}$.
I know $a|b \rightarrow b = aq, q \in \mathbb{Z}$ by the definition of divisibility.
I can then (I think) exponentiate both sides to n, since they are equivalent.
$b^n = (aq)^n$
$b^n = a^nq^n$
From here I would assume that I use induction of some sort to prove this generally holds for $n \in \mathbb{N}$
But how do I even go about doing that? If I make the induction assumption that for an arbitrary $k \in \mathbb{N}$ that $a^k|b^k$ does that not imply that $b^k = a^kq$ for some $q \in \mathbb{Z}$?
I feel like I've ran into a contradiction in my thought process, before even really getting to the $k+1$ induction step and I don't know how to proceed. I don't know how I'd even begin to show this holds by induction, since I am not even sure what induction assumption I should be making at this point.
Any hints or help would be greatly appreciated to get me unstuck.
Let there be some integer $p$ such that $$p\cdot a=b$$ You want to an integer $q$ such that $$q\cdot a^n=b^n$$
$$b^n=(p\cdot a)^n=p^na^n$$
You've just shown $b^n$ to equal some integer $p^n$ times $a^n$. Note that $p^n$ is an integer because an integer to the power of a natural number is an integer.
If you want to, you can show that an integer, $p$, raised to the power of a natural number $n$, is always going to be an integer using induction.
Base Case: $n=1$
$p^1=p$ which is an integer
Inductive Step:
Assume that $p^k$ is an integer for some natural number $k$.
$$p^{k+1}=p^k\cdot p^1$$
An integer, $p^k$, times integer $p$ is going to be an integer.