Question: > 
> Show that $(xy)^n=x^ny^n$ if $xy=yx$. 
> 
> 
> 

I assume I will need 3 different cases: $n < 0$, $n=0$, and $n > 0$.

For the $n > 0$ case, can I use induction? For the base case I'll show that if $n=$1 then $(xy)^1=x^1y^1$ which will give us xy=xy so the base case is satisfied. Then I'll assume that the statement is true for n+1. Is that on the right track? 

For $n=0$, is it just trivial? that $(xy)^0=x^0y^0$ and when talking about groups $x^0=1$ so we will have $1=$1 which is true. 

For $n < 0$, I was advised to use $(xy)^n=((xy)^{-1})^{-n}$. I'm not exactly sure how to go about this step. 

I don't think I completely understand the part where this only works when $xy=yx$. 

Any suggestions or help would be great! Thanks!

Answer: 
As John Habert pointed out the stipulation $xy = yx$ is required since $(xy)^2 = (xy)(xy)$ and you need to be able to commutate the two terms in the middle in order to say this equals $xxyy = x^2y^2$. 

You are pretty much on the right track I believe. I'll make my own humble attempt at a solution too then.

* *Case 1*: *$n \gt 0$*

Let us induct on $n$. The base case is obvious since $(xy)^1 = xy = x^1y^1$. Suppose the result is true for $n$. That is we assume $(xy)^n = x^ny^n$. Then $$(xy)^{n+1} = (xy)^n (xy) = (x^ny^n)(xy) = (x^ny^{n-1}y)(yx) = x^n . yyy..yxy$$

Since $xy = yx $, we may commutate the $x$ on the right with preceding $y$'s to equate this expression to;$$x^nxyyyy...y = x^{n+1}y^{n+1}$$

$\implies$ the result holds for $n+1$ and thus for all $n \in \Bbb N$. 

* *Case 2*: *$n \lt 0$*

As you have done $(xy)^n = ((xy)^{-1})^{-n} = ((yx)^{-1})^{-n} = (x^{-1}y^{-1})^{-n}$ according to the definition of negative exponents, due to commutativity and since $(yx)^{-1} = x^{-1}y^{-1}$. Now, $(-n) \gt 0$ for which we have proved the result. Therefore, 

$$(xy)^n = (x^{-1}y^{-1})^{-n} = (x^{-1})^{-n}(y^{-1})^{-n} = x^ny^n$$

This is again true due to the definition, $x^{-n} = (x^{-1})^{n}$

* *Case 3*: *$n = 0$*

This is trivially true as you have mentioned. 

I hope you understand how essential the condition $xy = yx$ is in this context. Which I believe was the only intention behind this exercise. Hope I helped.
It's simpler if you use two proofs by induction.

**Lemma.** If $xy=yx$, then $xy^n=y^nx$, for all $n\in\mathbb{N}$.

*Proof.* The base $n=0$ is obvious: $xy^0=x=y^0x$. Suppose the result holds for $n$; then
\begin{align}
xy^{n+1}&=xy^{n}y &&\text{(by definition)}\\
&=y^nxy &&\text{(by induction hypothesis)}\\
&=y^nyx &&\text{(by hypothesis)}\\
&=y^{n+1}x &&\text{(by definition)}
\end{align}
which ends the proof.

**Theorem.** If $xy=yx$, then $(xy)^n=x^ny^n$, for all $n\in\mathbb{N}$.

*Proof.* The base $n=0$ is obvious: $(xy)^0=1=x^0y^0$. Suppose the result holds for $n$; then
\begin{align}
(xy)^{n+1}&=(xy)^n(xy) &&\text{(by definition)}\\
&=(x^ny^n)(xy) &&\text{(induction hypothesis)}\\
&=x^n(xy^n)y \\
&=x^n(xy^n)y &&\text{(by the lemma)}\\
&=(x^nx)(y^ny)\\
&=x^{n+1}y^{n+1}
\end{align}
which ends the proof.

For $n<0$, set $g=x^{-1}$, $h=y^{-1}$ and $m=-n$; then
$$
gh=x^{-1}y^{-1}=(yx)^{-1}=(xy)^{-1}=hg
$$
so the theorem can be applied to $g$ and $h$. Now
$$
(xy)^n=((xy)^{-1})^m=(hg)^m=h^mg^n=y^nx^n.
$$