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?
 A: 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
$x^{n} < x^{n-1}y^{1} < x^{n-2}y^{2} < x^{n-3}y^{3} < ... < x^{2}y^{n-2} < x^{1}y^{n-1} < y^{n}$.$\blacksquare$
A: Hint: Use that
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b^1+\cdots ab^{n-1}+b^n)$$
A: The comments are trending pretty strongly that you should be using induction.  And, in one sense, they're not wrong.  Whenever you use the symbol "..." in the reasoning of a proof, it's usually a pretty blatant sign that you're trying to avoid induction.
But, you know, if this wasn't for a formal proof class, I'd give you high marks for this.  I understand your chain of reasoning well, and sometimes an intuitive argument can get lost in a sea of formality.  And I've never thought of doing it this way myself, so I'm glad you were able to convey your original thoughts effectively.  Gold star!
A: This is equivalent to
$$x>0,\\1<\frac yx\implies 1<\left(\frac yx\right)^n$$
or 
$$1<t\implies 1<t^n.$$
By induction,
$$1<t\implies 1<t^1$$
and
$$1<t,1<t^n\implies 1<t^{n+1}.$$
A: Induction is a workable approach here as mentioned by Matthew. Also you can use algebra as Graubner writes.

I would probably do it with calculus. Investigate the properties of the function $f(x) = x^n$ or maybe $f(x) = x^n - x^{n-k}$. Can you conclude where it is growing or where it could have zeroes?
