Given the usual field and ordering axioms for the real numbers, it isn't difficult to prove that $x<y$, without any further restrictions on the signs of $x$ and $y$, implies $x^n<y^n$, with $n$ an odd natural number. However, all the proofs I've seen and come up with myself seem to rely on distinguishing between the different possible combinations of signs of $x$ and $y$. I find such "case-by-case" proofs, especially of elementary facts, rather unappealing.
Now, my question is: does there exist an elegant one-line proof of the fact that $x<y\Rightarrow x^n<y^n$ (for $n$ an odd natural number) which doesn't take into account the signs of $x$ and $y$?
Thanks for reading and for any comments!