Edit: Wrong proof.

Notice that $y = (1+\frac{1}{n}) x = x+\frac{x}{n}$, so $x^y = x^{x + \frac{x}{n}} = x^{x} \cdot x^{\frac{x}{n}} = x^x \cdot (1+\frac{1}{n})^x$ and $y^x = [(1+\frac{1}{n})x]^x = x^x \cdot (1+\frac{1}{n})^x$. Hence, $\boxed{x^y = y^x}$.

modularmarc101's solution is correct.

Goutham wrote: I am sure this was posted somewhere in AoPS sometime ago but I can't find the link. \[(n+1)^2>n^2\Longrightarrow 1+\frac{1}{n}>\frac{n}{n+1}\] \[\Longrightarrow \frac{y}{x}>\log_xy=\frac{\log y}{\log x}\] \[\Longrightarrow y\log x>x\log y\] \[\Longrightarrow x^y>y^x\] You have a mistake: $\frac{n}{n+1} = \log_y x$, not $\log_x y$. Your proof actually reveals that $y^y > x^x$, not that $x^y > y^x$.

just a opinion question:How do mathmatican discover those intersting property?

that's what we can call the creativity sensitive.

modularmarc101 wrote: Notice that $y = (1+\frac{1}{n}) x = x+\frac{x}{n}$, so $x^y = x^{x + \frac{x}{n}} = x^{x} \cdot x^{\frac{x}{n}} = x^x \cdot (1+\frac{1}{n})^x$ and $y^x = [(1+\frac{1}{n})x]^x = x^x \cdot (1+\frac{1}{n})^x$. Hence, $\boxed{x^y = y^x}$. Also,can this be genralize to be a formula whereas in other case this still work?

Make the substitution $ t=n+ 1/n $ for simplicity and we have to compare two powers of $t$ , namely $n(t^n^+^1)$ and $(n+1)t^n.$ Dividing the former by the latter, we see that the quotient is $ (n/n+1)t = 1 $ (by our definition of t). Hence, they are both equal, and it follows that $ x^y = y^x $

$y=(1+ \frac{1}{n})^{n+1}=(1+ \frac{1}{n})x=(x+ \frac{x}{n})$ $x^y=x^{x+ \frac{x}{n}}=x^{x}x^{\frac{x}{n}}=x^x(1+ \frac{1}{n})^x$ $y^x=(x+ \frac{x}{n})^{(1+ \frac{1}{n})^{n}}=((1+ \frac{1}{n})x)^x=x^x(1+ \frac{1}{n})^x$ $\Rightarrow x^y=y^x$ $\blacksquare$