Is there a "one-line" proof of $xGiven 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!
 A: $f(t)=t^{2k+1}$ is a strictly monotonic function, as $f'(t)>0$ (except at $t=0$).
A: The "One Liner":
$$
x^n - y^n 
= \frac{(X-Y)^n - (X+Y)^n}{2^{n/2}} 
= \sum_{k=0}^n \binom{n}{k} \frac{\left[ X^{n-k}Y^k \left( (-1)^k - 1 \right) \right]}{2^{n/2}} 
= N Y 
= P (x - y)
$$
where $N$ is a negative and $P$ is a positive number (both dependent on the particular $x$, $y$).
Explanation:
\begin{align}
x^n - y^n 
&= \frac{1}{2^{n/2}} \left[ (X-Y)^n - (X+Y)^n \right] \\
&= \frac{1}{2^{n/2}} \sum_{k=0}^n \binom{n}{k} \left[ X^{n-k}(-1)^k Y^k - X^{n-k} Y^k \right] \\
&= \frac{1}{2^{n/2}} \sum_{k=0}^n \binom{n}{k} \left[ X^{n-k} Y^k ((-1)^k - 1) \right]  \quad (*) \\
&= N Y \\
&= \frac{N}{\sqrt{2}} (y - x) \\
&= -\frac{N}{\sqrt{2}}(x - y) \\
&= P (x - y)
\end{align}
This one uses a coordinate transformation $(x,y) \to (X,Y)$, a rotation by $-45^\circ$, to move all points with $x < y$ into the upper half plane such that the expression simplifies. 
Here are plots (3D, isolines) for $x^3 - y^3$ before and after the rotation:

$\quad$

$\quad \begin{matrix}\to \\ \\ \\ \\ \\ \end{matrix} \quad$

$\quad$

We see that $N$ consists mainly of the even powers of the binomial expansion, and then we transform back.
The necessary fact that $n$ is odd is used at $(*)$ to note that the differences $n-k$ are even, so $X$ shows up with even powers, while $Y$ stays with odd powers, so we can pull $Y$ out and are left with even powers of $X$ and $Y$.
A: $$\begin{align}y^n-x^n=(y-x)\sum_{k=0}^{n-1}x^ky^{n-1-k}=y^{n-1}(y-x)\underset{\text{$>1+x/y>0$ if} \ \lvert x/y\rvert<1}{\underbrace{\sum_{k=0}^{n-1}(x/y)^k}}&\ge  y^{n-1}(y-x)\underset{>0\ \text{if}\ \lvert x/y\rvert>1}{\underbrace{\sum_{k=0}^{(n-1)/2}\lvert x/y\rvert^{2k}(1-\lvert x/y\rvert^{-1})}}\end{align}$$
