Proof: If $xThe only thing I can come up with is this:
$$x<y$$
$$\Longleftrightarrow x\cdot x^{n-1}<y\cdot x^{n-1}$$
Surely $y\cdot x^{n-1}<y\cdot y^{n-1}$ because $x<y$
Then $x\cdot x^{n-1}<y\cdot x^{n-1}<y\cdot y^{n-1}$
$$x^n<y\cdot x^{n-1}<y^n,\text{ so }x^n<y^n$$
I feel this is wrong because I didn't use the fact that $n$ is odd anywhere, but it's all I could come up with.
 A: As was pointed out in the comments, the 3rd line of your attempt is wrong.
It's easy to prove this by examining the signs of $x$ and $y$:


*

*If $x\le0\le y$ (one of them sharp), then $x^n\le 0\le y^n$ (one of them sharp) and we are done.

*If $0\le x<y$, then $0\le\frac xy<1\implies (\frac xy)^n<1\implies x^n<y^n$.

*If $x<y\le 0$, then $1>\frac yx\ge 0\implies 1>(\frac yx)^n\implies x^n<y^n$, it gets reversed since $x^n<0$.


Alternatively, you are trying to prove $f(t) = t^n$ is an increasing function if $n$ is odd. We have derivates for this:
$$({t^n})'=nt^{n-1}=n(t^{(n-1)/2})^2\ge0$$
with equality only when $t=0$. Therefore the function is increasing on $\mathbb R\setminus\{0\}$ and hence necessarily on the whole $\mathbb R$.
A: Suppose for the sake of contradiction that there is a positive integer $k$ along with numbers $x,y$ with $x<y$ such that $x^{2k+1}-y^{2k+1}\geq 0$. It is well known (and a good inductive exercise as well) that $$x^{2k+1}-y^{2k+1} = (x-y)(x^{2k}+x^{2k-1}y+ \dots + xy^{2k-1}+y^{2k})$$ By assumption $x-y$ is a negative quantity, so we know $$(x-y)(x^{2k}+x^{2k-1}y+ \dots + xy^{2k-1}+y^{2k})\geq 0 \\ \implies x^{2k}+x^{2k-1}y+ \dots + xy^{2k-1}+y^{2k} \leq 0 $$ by division of $x-y$ to both sides. It should now be clear that if $x\geq 0$ and $y >0$ then the inequality is incorrect. So we must investigate the case of $x<0$ and $y \leq 0$. If $y=0$ the inequality will fail because everything cancels except for the term $x^{2k}>0$. So we now know $x<y<0$ is the last case we must look at. However, if both $x,y<0$ then $x^{2k}>0$, and $x^{2k-1}y>0$ because $x^{2k-1}$ and $y$ are negative. And $x^{2k-2}y^2>0$ because $x^{2k-2}$ and $y^2$ are positive. This argument holds for each term in the sum $$x^{2k}+x^{2k-1}y+ \dots + xy^{2k-1}+y^{2k}$$ because $x^{2k-i}y^{i}>0$ for all $i \in \{0,1,2,\dots, 2k\}$. So the sum has to be greater than zero, yielding your last contradiction.
A: The point of restricting $n$ to odd integers is that if $x<0$ and $n$ is odd then $x^n<0$ (this is not true for even $n$ since $x>0$ $\implies$ $x^n>0$ for even $n$). As a result, your problem becomes much simpler when considering different cases based on the sign of $x$ and $y$. Let's consider the case where $x<0<y$. Let $n$ be an odd integer. Since $x<0$ and $n$ is odd then $x^n<0$. Since $y>0$ then clearly $y^n>0$. Hence, $x^n<y^n$ and this completes the proof for the case $x<0<y$. 
I leave the cases of $0<x<y$ and $x<y<0$ to you (hint: use induction on $n$).
