Problem:
Prove that if $n < m$ with $n,m \in \mathbf{Z}$ and $x>1$, then $x^n < x^m$.
Proof:
Since $ x > 0$, we know $x^n > 0$.
Also, since $m > n$, it follows that $m - n \geq 1$. Thus $x^{m-n} > 1$.
Finally, $x^n(x^{m-n} -1) > 0$. Rearranging yields the desired result: $x^n < x^m$.
Question: I hope my proof is correct. I am actually curious whether it is possible to prove this result using induction? Could you somehow map $n,m$ from $\mathbf{Z}$ to the positive integers?
Induction proof(for positive $n$ and $m$):
As we have : $m > n > 0$. Let's prove the base : $m = 2, n = 1 : x^2-x > 0 $ - true for $x > 1$.
Now assume that's true for $n = 1$ and $m > n$, so let's prove it for $m = m+1$ :
$x^{m+1} - x > x^m - x > 0$ - true.
Now let's assume that's true for $n = n$ and $m > n$, let's prove it for $m = m+1$.
$x^{m+1}-x^n > x^m-x^n>0$ - true.
This proof could be expand for negative $m$ and $n$. Just use the same idea.
If $m =0$ or $n = 0$ then the proof reduced to compare $1$ and $x^{n}$ or $x^{m}$.