How to show that $(1+\frac1x)^x$ is increasing on $[0,+\infty[$ I am trying to show that the function $f(x)=(1+\frac1x)^x$ on $[0,+\infty[$.
I have found that $f'(x)=f(x) \left[\ln\left(\frac{x+1}x\right)- \frac1{x+1}\right]$.
Since $f(x)$ is always positive, I only have to show that $\ln\left(\frac{x+1}x\right)>\frac1{x+1}$ when $x>0$.
Is there an easy way to do this?
 A: A different approach altogether:
If we start with Bernoulli's inequality, $(1+u)^r\gt1+ru$ for $u\ge0$ and $r\ge1$ (which is easy to prove by taking the derivative of $f(u)=(1+u)^r-1-ru$), we have, on letting $u=rx$, 
$$\left(1+{1\over xr}\right)^r\ge1+r\cdot{1\over rx}=1+{1\over x}\implies\left(1+{1\over rx}\right)^{rx}=\left(\left(1+{1\over rx}\right)^r\right)^x\ge\left(1+{1\over x}\right)^x\quad\text{if }r\ge1$$
hence if $y=rx\ge x$ then
$$\left(1+{1\over y}\right)^y\ge\left(1+{1\over x}\right)^x$$
A: $$
\begin{align}
\log\left(1+\frac1x\right)
&=\int_0^{1/x}\frac{\mathrm{d}t}{1+t}\\
&\ge\int_0^{1/x}\frac{\mathrm{d}t}{1+1/x}\\[3pt]
&=\frac1{x+1}
\end{align}
$$
A: $$\ln\frac{x+1}x>\frac1{x+1}$$
$$\iff-\ln\left(1-\frac1{x+1}\right)>\frac1{x+1}$$
For $x>0$, $0<\frac1{x+1}<1$, so substitute $y=\frac1{x+1}$:
$$\iff-\ln(1-y)>y$$
The Maclaurin series of $\ln(1-y)$ is always valid for $0<y<1$:
$$\iff y+\frac{y^2}2+\frac{y^2}3+\dots>y$$
$$\iff\frac{y^2}2+\frac{y^3}3+\dots>0$$
which is true since $y$ is positive.
A: $$x>0\implies\log (1+1/x)=\log (1+x)-\log x=$$ $$=\int_x^{x+1}(1/t)dt>\int_x^{x+1}(1/(x+1)dt=1/(x+1).$$
A: Let
$g(x)
= \ln\left(\frac{x+1}x\right)-\frac1{x+1}
= \ln(x+1)-\ln(x)-\frac1{x+1}
$.
$\begin{array}\\
g'(x)
&=\frac1{x+1}-\frac1{x}+\frac1{(x+1)^2}\\
&=\frac{x(x+1)-(x+1)^2+x}{x(x+1)^2}\\
&=\frac{-1}{x(x+1)^2}\\
\end{array}
$
Therefore
$g(x)$ is decreasing.
But
$\lim_{x \to \infty} \ln(1+1/x)
= 0
$
and
$\lim_{x \to \infty} \frac1{1+x}
= 0
$,
so that
$\lim_{x \to \infty} g(x)
= 0
$.
Therefore
$g(x)> 0$
for $x > 0$.
A: This is an old problem and there are already clever solutions (see robjhon's solution above using integration)  and the comment by Paramanand Sing (using classical inequalities). Both strategies solve the problem under the assumption $x>0$, stated by the OP.
Here I just want to add that the statement is also valid for $x<-1$. The method follows the along the lines proposed by the OP, namely, to look at the sign of $f'(x)$. The whole method relies on a simple extension the inequalities mentioned by Paramanand Sing to all $x+1>0$.

Proposition: If $t+1>0$, then $$\frac{t}{1+t}\leq \log(1+t)\leq t$$
with equality iff $t=0$.

Proof: The right hand side follows immediately as $1+t\leq e^t$ for all $t\in\mathbb{R}$ with equality iff $t=0$. The left hand side can be obtained    from
$$\frac{1}{1+t}=1-\frac{t}{1+t}\leq \exp\big(-\frac{t}{1+t}\big)$$
by  taking logarithms on both sides of this inequality.
Using the computation of $f'(x)=f(x)\Big(\log\big(1+\tfrac{1}{x}\big)-\frac{1}{1+x}\Big)$ already provided by the OP, we obtain that for $x<-1$ or $x>0$, $1+\frac{1}{x}>0$ and so,
$$\log\big(1+\tfrac{1}{x}\big)-\frac{1}{1+x}=\log\big(1+\tfrac{1}{x}\big)-\frac{\tfrac1x}{1+\tfrac1x}>0$$
Putting things together, we obtain that $f$ is monotone increasing in $(-\infty,-1)$, and also monotone increasing in $(0,\infty)$.

Comment: The same technique works for showing that $g:x\mapsto \big(1+\frac{1}{x}\big)^{x+1}$ is monotone decreasing:
for all $x<-1$ or $x>0$
$$g'(x)=g(x)\Big( \log(1+\tfrac1x)-\tfrac{1}{x}\Big)<0$$
which follows by the proposition above.
All these yield following useful inequality
$$\Big(1+\frac{1}{x}\Big)^x<e<\Big(1+\frac{1}{x}\Big)^{x+1}$$
for all $x>0$ and
$$\Big(1+\frac{1}{x}\Big)^{x+1}<e<\Big(1+\frac{1}{x}\Big)^x$$
for all $x<-1$.
