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? 
 A: No AM-GM inequality - just simple computation: 
$$\begin{align}
\frac{(1+\frac{x}{n+1})^{n+1}}{(1+\frac{x}{n})^n} 
  &= (1+\frac{x}{n})\left(\frac{1+\frac{x}{n+1}}{1+\frac{x}{n}}\right)^{n+1} \\\\
  &= (1+\frac{x}{n})\left(\frac{n(n+1)+nx}{(n+1)(n+x)}\right)^{n+1} \\\\
  &= (1+\frac{x}{n})\left(\frac{(n+1)(n+x)-x}{(n+1)(n+x)}\right)^{n+1} \\\\
  &= (1+\frac{x}{n})\left(1-\frac{x}{(n+1)(n+x)}\right)^{n+1} \\\\
  &> (1+\frac{x}{n})(1-\frac{x}{n+x}) = \frac{n+x}{n} \frac{n}{n+x} = 1.
\end{align}$$
Copied from a previous answer of mine.
A: 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 
$$
\left( 1+ \frac{1}{n+1} \right)^{n+1} \gt \left( 1 + \frac{1}{n} \right)^n.
$$
A: 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 $\left(1+\frac1n\right)^n$.
A: Here's a direct argument without using AM-GM: write
$$\left(1+{1\over n}\right)^n=\sum_{j\geq 0} {n\choose j}\left({1\over n}\right)^j=\sum_{j\geq 0}\,\, \prod_{0\leq k<j}\left(1-{k\over n}\right) \cdot{1\over j!}.$$ 
Each product inside the sum gets bigger as $n$ increases, and so the same is true for whole sum. 
A: As requested, here is a proof using Bernoulli's inequality.
$(1+x)^r \ge 1 + rx$, for any real $x \gt -1$ and real $r \ge 1$.
We set $r = \frac{n+1}{n}$ and $x = \frac{1}{n+1}$.
We get
$$ \left(1 + \frac{1}{n+1}\right)^{(n+1)/n} \ge 1 + \frac{1}{n}$$
Taking $n^{th}$ power on both sides gives us the inequality.
$$ \left(1 + \frac{1}{n+1}\right)^{n+1} \ge \left(1 + \frac{1}{n}\right)^n$$
Now we only need to eliminate the equality portion.
Assume they were equal, then we must have that
$$(n+2)^{n+1}n^n = (n+1)^{2n+1}$$
which is not possible as $n+1$ is relatively prime with both $n$ and $n+2$. (Of course, we could probably have used a strict version of Bernoulli's inequality...).
A: The calculus argument: taking logarithms of $(1+1/n)^n$, it's enough to show that $f(x) = x \log (1+1/x)$ is an increasing function of $x$ for $x > 0$. Now 
$$ f^\prime(x) = \log \left( 1 + {1 \over x} \right) - {1 \over x+1} $$
and it suffices to show this is positive. So we need $\log (1 + 1/x) > 1/(x+1)$; taking exponentials it suffices to show that $1 + {1 \over x} > \exp \left( {1 \over x+1} \right)$ when $x > 0$. But we have
$$ \exp(z) = 1 + z + {z^2 \over 2!} + {z^3 \over 3!} + \cdots < 1 + z + z^2 + z^3 + \cdots = {1 \over 1-z} $$
whenever $|z|<1$. Letting $z = 1/(x+1)$ gives $e^{1/(x+1)} < 1 + 1/x$, as desired.
