Proving : $ \bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n} $ 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 equivalent to showing that
$$
\left(\frac{n}{n-1}\right)^{n-1}\tag{1}
$$
is an increasing function of $n$. Consider the Taylor expansion of
$$
\begin{align}
(n-1)\log\left(\frac{1}{1-1/n}\right)
&=(n-1)\left(\frac1n+\frac12\frac1{n^2}+\frac13\frac1{n^3}+\dots\right)\\
&=1-\frac1{1\cdot2}\frac1n-\frac1{2\cdot3}\frac1{n^2}-\frac1{3\cdot4}\frac1{n^3}+\dots\tag{2}
\end{align}
$$
$(2)$ is obviously an increasing function of $n$. QED
Another useful case
$$
\left(\frac{n}{n-1}\right)^n\tag{3}
$$
is a decreasing function of $n$.
$$
\begin{align}
n\log\left(\frac{1}{1-1/n}\right)
&=n\left(\frac1n+\frac12\frac1{n^2}+\frac13\frac1{n^3}+\dots\right)\\
&=1+\frac12\frac1n+\frac13\frac1{n^2}+\dots\tag{4}
\end{align}
$$
$(4)$ is obviously a decreasing function of $n$.
A boundary case
$$
\left(\frac{n}{n-1}\right)^{n-1/2}\tag{5}
$$
is a decreasing function of $n$.
$$
\begin{align}
(n-1/2)\log\left(\frac{1}{1-1/n}\right)
&=(n-1/2)\left(\frac1n+\frac12\frac1{n^2}+\frac13\frac1{n^3}+\dots\right)\\
&=1+\frac12\left(\frac1{2\cdot3}\frac1{n^2}+\frac2{3\cdot4}\frac1{n^3}+\frac3{4\cdot5}\frac1{n^4}+\dots\right)\tag{6}
\end{align}
$$
$(6)$ is obviously a decreasing function of $n$.
Comparing $(6)$ to $(2)$ shows that $\left(1+\frac1n\right)^{n+1/2}$ is a lot closer to $e$ than is $\left(1+\frac1n\right)^n$.
