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?

$$ \begin{align} \log\left(1+\frac1x\right) &=\int_0^{1/x}\frac{\mathrm{d}t}{1+t}\\ &\ge\int_0^{1/x}\frac{\mathrm{d}t}{1+1/x}\\[3pt] &=\frac1{x+1} \end{align} $$

A 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

$$\left(1+{1\over y}\right)^y\ge\left(1+{1\over x}\right)^x$$

$$\ln\frac{x+1}x>\frac1{x+1}$$ $$\iff-\ln\left(1-\frac1{x+1}\right)>\frac1{x+1}$$ 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.