The logarithm lies below its tangent

Let $f(x)=x-\ln(1+x)$ for $x>-1$. Then
$$f'(x)=1-\frac1{1+x}=\frac{x}{1+x}.$$

Since $1+x>0$, $f'(x)<0$ on $(-1,0)$ and $f'(x)>0$ on $(0,+\infty)$. So $f$ has its minimum at $0$, where $f(0)=0$. Hence $f(x)\ge0$, that is $\ln(1+x)\le x$.