$|e^a-e^b| \leq |a-b|$ for complex numbers with non-positive real parts

Came across this problem on an old qualifying exam: Let $a$ and $b$ be complex numbers whose real parts are negative or 0. Prove the inequality $|e^a-e^b| \leq |a-b|$.

If $f(z)=e^z$ and $z=x+iy$, then $|f'(z)|=e^x\leq 1$ given that $x \leq 0$. I played around with the limit definition of the derivative, but wasn't able to get anywhere. Not sure what else to try; a hint would be very helpful!

Consider integrating $f'(z) dz$ along the line segment from $a$ to $b$

Prove and then use the following fact:

Let $D \subseteq \mathbb C$ be a convex region and let $f: D \to \mathbb C$ be holomorphic with $|f'|\le 1$ on $D$. Then for $a,b\in D$ we have

$$ |f(b) - f(a)| \le |b-a|$$

An interesting related article A norm inequality for Hermitian operators by Ritsuo Nakamoto

The American Mathematical Monthly; Mar 2003; 110, 3;