A while ago I posted an attempt at a proof of $x<y \iff x^n<y^n$. It was pointed out that I hadn't actually used induction, and had instead done a direct proof. Below is the link to the question, so please do not mark this question as a duplicate, as this new question is about whether I have now done the proof by induction correctly, rather than accidentally reverting to a direct proof.
Also, be aware that I am, below, attempting to prove only that $x<y \implies x^n<y^n$.
Claim: $x<y \implies x^n<y^n$ for $x,y>0$ and $x,y,n \in \mathbb N$.
Proof:
Let $P(n)$ be the statement that $$x<y \implies x^n<y^n,$$ for $n\in \mathbb N$. It is clear that $P(n)$ holds for $n=1$ since $x<y \implies x<y$.
Assuming that $P(n)$ holds for some $n = k$, we see that this implies that $P(n)$ holds for $n=k+1$, as follows.
$$x<y \implies x^n<y^n$$
Since we know that $x<y$, if we multiply $x^n<y^n$ by $x$ we get that: $$x^{n+1} < xy^n,$$ from which it follows that $$x^{n+1} < y^{n+1}.$$
Thus $P(k)$ true $\implies$ $P(k+1)$ true, and so by induction we can prove the claim that $P(n)$ holds for all $n \in \mathbb N$.
Best Answer
In my opinion you should work better out where and how you use the inductive claim (I. C.).
$x^{n+1}=x\cdot x^n\stackrel{I.C}{<}x\cdot y^n\stackrel{x<y}{<}y\cdot y^n=y^{n+1}$