How could we prove that this inequality holds 


  $$ \left(1+\frac{1}{n+1}\right)^{n+1} \gt \left(1+\frac{1}{n} \right)^{n} $$


where $n \in \mathbb{N}$, I think we could use the AM-GM inequality for this but not getting how? 

This is one of the cutest applications of AM-GM I have learned. Unfortunately, I do not remember the source. 

Define the numbers $x_0, x_1, x_2, \ldots, x_n$ by:
$$
x_i =
\begin{cases}
1, &i = 0,
\\\\ 1+\frac{1}{n}, &1 \leqslant i \leqslant n. 
\end{cases}
$$
The claim follows by applying AM-GM:
$$
\left( \frac{x_0 + x_1 + \ldots + x_n}{n+1} \right)^{n+1} \gt \ \prod_{i=0}^n \, x_i .
$$
Plugging in the above values, we get 
$$
\left( \frac{1+n \Big(1+\frac{1}{n} \Big)}{n+1} \right)^{n+1} \gt \  1 \cdot \left( 1+\frac{1}{n} \right)^n ,
$$
which simplifies to 
$$
\left( 1+ \frac{1}{n+1} \right)^{n+1} \gt \left( 1 + \frac{1}{n} \right)^n.
$$
