Claim: $x<y \iff x^n < y^n$ for $n\in \mathbb N$. Edit: $x,y>0$
Proof:
Since that which is to be proved is biconditional we must prove both that $x<y \implies x^n<y^n$ and $x<y \impliedby x^n<y^n$ are true.
First we prove that $x<y \implies x^n<y^n$ is true by induction. It is trivial to show that this is true for the case $n=1$, and I have proved earlier that it is also true for the case $n=2$ (so we do not need to prove this here.
Let us assume that the statement holds for some $n=k$. We seek to show that it is also true for $n=k+1$.
We can say that $$x^{k+1} – y^{k+1} \equiv (x-y)(x^k+x^{k-1}y +…+xy^{k-1} + y^k)$$ and since $(x-y)<0$, and $(x^k+x^{k-1}y +…+xy^{k-1} + y^k) > 0$, we see that $x^{k+1} – y^{k+1}$ must be negative. Therefore we arrive at the desired result:
$$x^{k+1} – y^{k+1}<0.$$
Thus for all $n\in \mathbb N$, $x<y \implies x^n<y^n$.
Now we must prove that $x<y \impliedby x^n<y^n$ is also true. If we write this statement in the contrapositive form, we find that:
$$(x<y \impliedby x^n<y^n) \iff (x^n<y^n \implies x<y)$$
$$\iff (y<x \implies y^n<x^n).$$
Since $x$ and $y$ are arbitrary numbers, we have already proven this. $$\tag*{$\blacksquare$}$$
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The above is my attempt at the proof, I would like for someone to confirm whether it is correct or not. For the line where I state the equivalence $x^{k+1} – y^{k+1} \equiv (x-y)(x^k+x^{k-1}y +…+xy^{k-1} + y^k)$, I must make clear that I have had to use this as a given and would appreciate if someone could point out why this is obvious or how to go about proving this, almost as a lemma for the proof. My final request is that if there is any problem with my actual proofwriting, I ask that you point it out (e.g is this a usual style, or is it unsual and hard to follow, etc).
Edit: It was pointed out that I forgot to include that $x$ and $y$ are positive, so I have included this.
Best Answer
Hint:
For $n=1$, $x<y\iff x<y$ obviously holds.
Now assume that for some $n$,
$$x<y\iff x^n<y^n.$$
Then by the rule of multiplication of inequalities,
$$x<y\land x^n<y^n\implies x^{n+1}<y^{n+1}$$
so that
$$x<y\implies x^n<y^n\implies x^{n+1}<y^{n+1}.$$
Now try the contrapositive.