How to prove $\left(1+\frac{x}{n+1}\right)^{n+1} > \left(1+\frac{x}{n}\right)^n$ for all $n \in \mathbb Z^+$ and $x>0$ How to prove $\left(1+\frac{x}{n+1}\right)^{n+1} > \left(1+\frac{x}{n}\right)^n$ for all $n \in \mathbb Z^+$ and $x>0$
I tried by using AM-GM inequality. 
Can anyone give me some help.

To show that using AM-GM, consider
$$a_1=1, a_2 = a_3 = \cdots a_{n+1} = 1+\frac{x}n$$
Now AM-GM gives the strict inequality:
$$1+\frac{x}{n+1} = \frac{1+n(1+\frac{x}n)}{n+1}=\frac{\sum a_i}{n+1} > \sqrt[n+1]{\prod a_i} = \left(1+\frac{x}n \right)^{\frac{n}{n+1}}$$
