If $x>y>0$ , then $\frac{x^y y^x}{y^y x^x}=$

$\text{(A) } (x-y)^{y/x}\quad \text{(B) } \left(\frac{x}{y}\right)^{x-y}\quad \text{(C) } 1\quad \text{(D) } \left(\frac{x}{y}\right)^{y-x}\quad \text{(E) } (x-y)^{x/y}$
We see that this fraction can easily be factored as $\frac{x^y}{y^y}\times\frac{y^x}{x^x}$. Since $\frac{y^x}{x^x}=\frac{x^{-x}}{y^{-x}}$, this fraction is equivalent to $\left(\frac{x}{y}\right)^y\times\left(\frac{x}{y}\right)^{-x}=\left(\frac{x}{y}\right)^{y-x}\quad$, which corresponds to answer choice $\fbox{D}$.