Want to ensure my proof is rigourous enough. Question.
Prove: $ 0 \leq x < y $ then $ x^n < y^n$ $ \forall n \in \mathbb{N} $
I'm particularly bad at proving obvious things but here it goes. ( please be super strict on analyzing my proof please.)
Im not sure how to do this one algebraically, my first thought is i want to prove something a bit different i want to prove that  $ \dfrac {x} {y} <1$ then apply that to $ (\dfrac {x} {y})^n < 1 $ 
im lazy so lets call  $ \dfrac {x} {y} = r$
since $ 0 \leq r < 1$ i want to show that $ r^n < 1$ $ \forall n \in \mathbb{N}$ i want to use some kind of geometric principle or well really anything to get me off the hook of having to use induction at this point. would showing that $ \lim_{n \to \infty} r^n \to 0$ do anything for me?
if not i guess $r_1 < 1$ as x < y
$ r_2 < r_1 $ as $ r_2 = r_1 * r_1 $ hence $ \dfrac {r_1^2} { r_1} < \dfrac {r_1} { r_1}$ or $r_1 < 1$ which we already established
then  we want to show that $r_{n-1} > r_n $ for all n.
first we should note that we know $r_n = r_1 * r_{n-1}$
This leads us to  = $r_{n-1} > r_{n-1} * r_1 $  which is easy as $( \dfrac {r_{n-1}} {r_{n-1}} > \dfrac {r_{n-1} * r_1} {r_{n-1}} )$ = $ 1> {r_{1}}$ which we already know.
i suck at induction so it feels like i am missing something here to proving it this way.
 A: I think you mean $0 \leqslant x < y$ in your first line. As an alternative method, consider the following.
The base case for your induction ($n=1$) is of course true as $x<y$ and you've already stipulated that $0 \leqslant x < y$.
Assumption Step
Assume that the inequality is true for an arbitrary natural number $k$, that is to say that $x^k < y^k$.
Inductive Step
Multiplying this inequality by $x$ gives $$ x^{k+1} < y^k x $$ and multiplying it by $y$ gives $$x^k y < y^{k+1}$$
Since $0 \leqslant x < y$, the directions of the inequalities don't change and it's clear to see that $x^{k} y < y^{k} x$. Using the transitivity of inequalities, what can you conclude?
A: Another way to complete the induction is by contradiction. Assume for all $k<n$ that $x^k < y^k$. Now assume that $x^n \geq y^n$. But this would yield $$\frac{y^n}{x}\geq x^{n-1}\Rightarrow (\frac{y}{x})y^{n-1}\geq x^{n-1}$$
Use the fact that $\frac{y}{x} > 1$ to get a contradiction with the induction  hypothesis.
A: The idea is good.
First of all, if $x=0$ there's nothing to prove, except that $y^n>0$ for all $n$.
So let $x>0$ and set $r=x/y<1$ and try proving that $r^n<1$. For $n=1$ this is obvious, so assume $r^n<1$. Then, multiplying by $r$ you get
$$
r^{n+1}=r^nr<1\cdot r<1
$$
Note: this uses the known fact that $0<a<b$ implies $0<ac<bc$ for all $c>0$.
