Show that for all $z \in \overline{D}(0;1)$, $(3-e)|z| \leq |e^z - 1|\leq |z|(e-1)$ 
Show that for all $z \in \overline{D}(0;1)$, $(3-e)|z| \leq |e^z - 1|\leq |z|(e-1)$

I think I'm supposed to use the following chain of inequalities
$$|e^z -1|\leq e^{|z|}-1 \leq |z|e^{|z|}$$
But every time I try to solve it I get stuck because some term doesn't vanish or I get something similar but no the same I'm asked to prove. Any help or hint will be appreciate and thanks in advance!
 A: Start with the series expansion 
$$e^z - 1 = z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots$$
and use the triangle and reverse triangle inequalities to get the result.

By the triangle inequality and the condition $\lvert z\rvert \le 1$, we have\begin{align}\lvert e^z - 1\rvert &\le \lvert z \rvert + \frac{\lvert z\rvert^2}{2!} + \frac{\lvert z\rvert^3}{3!} + \cdots\\ &\le \lvert z\rvert + \frac{\lvert z\rvert}{2!} + \frac{\lvert z\rvert}{3!} + \cdots = \lvert z\rvert\left(1 + \frac{1}{2!} + \frac{1}{3!} + \cdots\right) = \lvert z\rvert(e - 1)\end{align}
  On the other hand, by the reverse triangle inequality,\begin{align}\lvert e^z - 1\rvert &\ge \lvert z\rvert - \left\lvert\frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\right\rvert \ge \lvert z\rvert - \frac{\lvert z\rvert^2}{2!} - \frac{\lvert z\rvert^3}{3!} -\cdots \\
&\ge \lvert z\rvert - \frac{\lvert z\rvert}{2!} - \frac{\lvert z\rvert}{3!} - \cdots = \lvert z\rvert\left[1 - \left(\frac{1}{2!} + \frac{1}{3!} + \cdots\right)\right] = \lvert z\rvert(1 - (e - 2)) = \lvert z\rvert(3 - e)\end{align}Hence, $(3 - e)\lvert z\rvert \le \lvert e^z - 1\rvert \le (e - 1)\lvert z\rvert$ on $\overline{D}(0;1)$.

