Using binomial Theorem how we can show $\frac{(x+y)!}{x!y!}\leq \frac{(x+y)^{x+y}}{x^xy^y}$? Using binomial Theorem prove that $$\frac{(x+y)!}{x!y!}\leq \frac{(x+y)^{x+y}}{x^xy^y}.$$
I tried it as follows:
 It is clear that $x\leq x+y, \forall x,y\in \mathbb{N}$. Thus, by Binomial Theorem, we have
\begin{align*}(x+y)^{x+y}&=\displaystyle\sum_{y=0}^{x+y}{x+y\choose y}x^{(x+y)-y}y^{y}\\&=\sum_{y=0}^{x+y}\frac{(x+y)!}{x!y!}x^{x}y^{y}\\&\geq \frac{(x+y)!}{x!y!}x^{x}y^{y}(how?)\end{align*}
I can't show the last inequality, thus is there any one who can give me hint over here, please? Thanks . 

Hint: Expand $(x+y)^{x+y}$ and look at the $(y+1)^{th}$ term. Does it equal your LHS?
