Is this proof of $x

Claim: $xEdit: $x,y>0$

Proof:
Since that which is to be proved is biconditional we must prove both that $x

First we prove that $x

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

Now we must prove that $x

$$(x$$\iff (y

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.

Answer

Hint:

For $n=1$, $x

Now assume that for some $n$,

$$x

Then by the rule of multiplication of inequalities,

$$x

so that

$$x

Now try the contrapositive.