$ \left( 1 - \frac{x}{n}\right)^n \leq \left( 1 - \frac{x}{n+1}\right)^{n+1} $, $\forall$ $x$ $\in$ $[0,n]$ 
Prove that:
  $$ \left( 1 - \frac{x}{n}\right)^n \leq \left( 1 - \frac{x}{n+1}\right)^{n+1} $$ $\forall$ $x$ $\in$ $[0,n].$

I think it's easy, but I'm very stuck in this question, I need to prove that
$$\left(1- \frac{x}{n} \right)^n \leq \left(1- \frac{x}{n+1} \right)^{n+1}, \hspace{0.1cm} \forall \hspace{0.1cm} x \in [0,n].   $$
Any help?
 A: Apply the AM GM inequality to the $n+1$ numbers
$$ 1, 1-\frac{x}{n}, 1-\frac{x}{n}, \ldots, 1-\frac{x}{n}$$
The sum is $n+1-x$ and product is $\left(1-\frac{x}{n}\right)^n$. Hence
$$\left(1-\frac{x}{n}\right)^{\frac{n}{n+1}} \leq \frac{n+1-x}{n+1} = 1-\frac{x}{n+1}$$
Raise the sides to the power $n+1$.
A: $$\left(1-\frac{x}{n+1}\right)^{n+1}=\left(1-\frac{x}{n}+\frac{x}{n}-\frac{x}{n+1}\right)^{n+1}=$$
$$=\left(1-\frac{x}{n}+\frac{x}{n(n+1)}\right)^{n+1}=\left(1-\frac{x}{n}+\frac{x}{n(n+1)}\right)^n\left(1-\frac{x}{n}+\frac{x}{n(n+1)}\right)\geq$$
$$\geq\left(\left(1-\frac{x}{n}\right)^n+n\left(1-\frac{x}{n}\right)^{n-1}\cdot\frac{x}{n(n+1)}\right)\left(1-\frac{x}{n}+\frac{x}{n(n+1)}\right).$$
Thus, it remains to prove that
$$\left(\left(1-\frac{x}{n}\right)^n+\left(1-\frac{x}{n}\right)^{n-1}\cdot\frac{x}{n+1}\right)\left(1-\frac{x}{n+1}\right)\geq\left(1-\frac{x}{n}\right)^n,$$
which is
$$\frac{x^2}{n(n+1)^2}\left(1-\frac{x}{n}\right)^{n-1}\geq0.$$
Done!
A: Here is a direct solution without using any tricky manipulation. The inequality holds if $x=n$ and hence let $n>x\geq 0$. By the general binomial theorem we have $$\left(1-\frac{x}{n}\right)^{-n}=1+x+\dfrac{1+\dfrac{1}{n}}{2!}x^{2}+\dfrac{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}{3!}x^{3}+\dots$$ and the expression on right is non-increasing as $n$ increases. This proves the result in question (via taking reciprocals). 
