How do I prove $\cfrac{x^m}{x^n}=x^{m-n}$ for $x>0$ and $m, n \in \mathbb{N}$ with $m>n$
For me, I started with defining exponentiation, $$x^r=\underbrace{x \times x \times \ ... \times x}_{\text{r times}}$$ And so I say $$\cfrac{\underbrace{x \times x \times \ ... \times x}_{\text{m times}}}{\underbrace{x \times x \times \ ... \times x}_{\text{n times}}}\implies x^{m-n}$$
I think this is clear because of the fact that equal products in the numerator and denominator will cancel each other out until there's nothing left to cancel out, hence $m-n$. But I'm not sure if this is a good or solid approach.
Even if this is all the explanation I need, are there any interesting methods to prove this?
I suppose you are assuming that $x>0$, $n$ and $m$ are positive integers and $m \geq n$. What you want to prove is equivalent to $x^{m}=x^{n}x^{m-n}$. So this follows from the fact that $x^{i+j}=x^{i}x^{j}$ (with $i=n, j=m-n$).