For $x+y=n$, $y^x < x^y$ if $x(updated)
I'd like to use this property for my research, but it's somewhat messy to prove. 
$$\text{For all natural number $x,y$ such that $x+y=n$ and $1<x<y<n$, then $y^x < x^y$}.$$
For example, let $x=3, y=7$. Then $y^x = y^3 = 343$ and $x^y = 3^7 = 2187$. Any suggestion on how to prove this?

I proved this in the special case $x = 99, y = 100$, here.  As others have pointed out, what you really want to hold is the following: 
Statement: Let $x, y \in \mathbb{R}$.  Then $y > x > e$ implies $x^y > y^x$.
Proof:.  Write $y = x + z$, where $z > 0$.  Then,
$$\begin{align}
x^y > y^x &\iff  x^x x^z > y^x
\\
&\iff x^z > \left(\frac{x+z}{x} \right)^x
\\
&\iff x^z > \left( 1 + \frac{z}{x} \right)^x.
\end{align}$$
The right hand side $\left(1 + \frac{z}{x} \right)^x$ is monotone increasing with limit $e^z$.  Since the left hand size is strictly greater than $e^z$ (as $x > e$), it follows that the inequality always holds.
