<s>[INST] Proof: For all n > 0, if $0I need to solve this problem without using induction, only with the following properties.
(1) Trichotomy law; 
(2) if $a$ and $b$ are in $P$, then $a+b$ is in $P$; 
(3) if $a$ and $b$ are in $P$, then $ab$ is in $P$.
Where P is the collection of all Positive Numbers.
My idea is to use the fact that $y^n - x^n = (y - x)(x^{n-1} + yx^{n-2} + ... + xy^{n-2} + y^{n-1})$.
Thus, since $x \in P$, and $y \in P$, I can build, by (2) and (3), the expansion ($x^{n-1} + yx^{n-2} + ... + xy^{n-2} + y^{n-1})\in P$.  How $(y - x)\in P$ (Since x < y), and by (3), $(y - x)(x^{n-1} + yx^{n-2} + ... + xy^{n-2} + y^{n-1}) \in P$, which implies that $y^n - x^n\in P$, so $x^n < y^n$.
I'm in the right way? Thanks in advance.
 [/INST] First a comment:
You don't need the assumption that
$x^2 < y^2$.
In general,
any statement of the form
$\forall n P(n)$
can only be proved by induction,
ultimately.
In your case,
$y^n - x^n = (y - x)(x^{n-1} + yx^{n-2} + ... + xy^{n-2} + y^{n-1})
$
has to be proved by induction
with $P(n)$ being
$y^n-x^n
=(y-x)\sum_{k=0}^{n-1} x^ky^{n-1-k}
$.
Another proof on the factorization
 could use
$\begin{array}\\
(y+x)(y^n-x^n)
&=y^{n+1}-yx^n+xy^n-x^{n+1}\\
&=y^{n+1}-x^{n+1}+xy(y^{n-1}-x^{n-1})\\
\end{array}
$
or
$y^{n+1}-x^{n+1}
=(y+x)(y^n-x^n)-xy(y^{n-1}-x^{n-1})
$.
From this,
if
$x^n-y^n = (x-y)q_n(x, y)$
for $n$ and $n-1$
then
$\begin{array}\\
y^{n+1}-x^{n+1}
&=(y+x)(y-x)q_n(y, x)-xy(y-x)q_{n-1}(y, x)\\
&=(y-x)((y+x)q_n(y, x)-xyq_{n-1}(y, x))\\
\end{array}
$
so
$q_{n+1}(y, x)
=(y+x)q_n(y, x)-xyq_{n-1}(y, x)
$.
</s>