<s>[INST] If $0 < x < y$, then $x^n < y^n$ The problem asks to prove that if $0 < x < y$ then $x^{n} < y^{n}$, where $n$ is a positive integer, so I started by assuming that $0 < x < y$. I then wrote this chain of inequalities: $x^{n} < x^{n-1}y^{1} < x^{n-2}y^{2} < x^{n-3}y^{3} < ... < x^{2}y^{n-2} < x^{1}y^{n-1} < y^{n}$. It is true that $x^{n} < x^{n-1}y^{1}$, since dividing both sides by $x^{n-1}y^{0}$ yields $x < y$. Similarly, dividing both sides of $x^{n-1}y^{1} < x^{n-2}y^{2}$ by $x^{n-2}y^{1}$ yields $x < y$. It seems to me that this pattern continues for all the inequalities, but I haven't been able to prove this. Does that make the proof invalid? Also, how can I prove that this pattern always holds true?
 [/INST] Your proof is absolutely correct. But to make the proof clearer and standard I would recommend you to write these first.
$\because$ $0<x<y$.
$\implies \frac{y}{x}>1$  .....$(i)$
Multiplying both sides by $x^n$ we get,
$x^{n-1}y^1>x^n$
Again Multiplying both sides of $(i)$ by $x^{n-1}y$ we get,
$x^{n-2}y^2>x^{n-1}y$
Continuing this process of multiplying LHS of the obtained inequality to $(i)$, we finally obtain,
$y^n>x^1y^{n-1}$
Now you can write
$x^{n} < x^{n-1}y^{1} < x^{n-2}y^{2} < x^{n-3}y^{3} < ... < x^{2}y^{n-2} < x^{1}y^{n-1} < y^{n}$.$\blacksquare$
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