I think I am almost there:
Prove $\left|z\right|/4 < \left|\exp(z)-1\right|<7\left|z\right|/4$ for all $0<|z|<1$.
MY ADVANCES First we note that $$ \left|\exp(z)-1\right| = \left|\sum_{n=1}^\infty \frac{z^n}{n!}\right|= \left|z+\sum_{n=2}^\infty \frac{z^n}{n!}\right| $$ Then by using both sides of the triangle inequality we get $$ \left|z\right|-\sum_{n=2}^\infty \frac{\left|z\right|^n}{n!} \leq \left|\exp(z)-1\right|\leq \left|z\right|+\sum_{n=2}^\infty \frac{\left|z\right|^n}{n!}. $$ Then we only need to prove that if $0<\left|z\right|<1$ then $$ \sum_{n=2}^\infty \frac{\left|z\right|^n}{n!} < \frac{3}{4}\left|z\right| \ \ \ \ \ \ \ \ \ (1) $$ Where I am Stuck But here is as far as I have come, I can't get this last inequality, since I only get that $$ \sum_{n=2}^\infty \frac{\left|z\right|^n}{n!} < e-2 $$ Any hint to prove (1)? Is this the way on proving this?
For $0 <|z| < 1$,
$$\sum_{n = 2}^\infty \frac{|z|^n}{n!} \le \sum_{n = 2}^\infty \frac{|z|}{n!} = (e - 2)|z| < \frac{3}{4}|z|.$$