The exponential lies above its tangent

Let $f(x)=e^x-1-x$. Then $f'(x)=e^x-1$.

• For $x<0$ we have $e^x<1$, so $f'(x)<0$ and $f$ is strictly decreasing on $(-\infty,0]$.
• For $x>0$ we have $e^x>1$, so $f'(x)>0$ and $f$ is strictly increasing on $[0,+\infty)$.

So $f$ has a strict minimum at $0$, where $f(0)=0$. Therefore $f(x)>0$ for $x\ne0$, which gives $e^x\ge1+x$ with equality only at $x=0$.