show that if $n\geq1$, $(1+{1\over n})^n<(1+{1\over n+1})^{n+1}$
Solution 1:
use $AM-GM$,$$(1+\dfrac{1}{n})(1+\dfrac{1}{n})\cdots(1+\dfrac{1}{n})\cdot 1\le\left(\dfrac{n+1+1}{n+1}\right)^{n+1}$$
Solution 2:
$$\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} = 1-\frac{1}{(n+1)^2} $$
By Bernoulli's inequality:
$$\left(1-\frac{1}{(n+1)^2}\right)^n \geq 1-\frac{n}{(n+1)^2} $$
Putting it together:
$$\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^n} \geq \left(1-\frac{n}{(n+1)^2}\right)\left(1+\frac{1}{n+1}\right) = 1+\frac{1}{(n+1)^3} $$