<s>[INST] How to prove that for all $|z|<1$, $|(1-z)e^z-1|\leq |z|^2$? 
For all $|z|<1$, $$|(1-z)e^z-1|\leq |z|^2.$$


Applying Taylor's expansion (does it hold for all $z$?), we have 
$$\begin{aligned}|(1-z)e^z-1|&=\left|(1-z)\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots \right)-1~\right|\\&=\left|\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots \right)-\left(z+z^2+\frac{z^3}{2!}+\frac{z^4}{3!}+\cdots\right)-1~\right|\\
&=\left|~\left(\frac{1}{2!}-\frac{1}{1!}\right)z^2+\left(\frac{1}{3!}-\frac{1}{2!}\right)z^3+\cdots~\right|\end{aligned}$$
How should I go on? 
 [/INST] In this particular case, one does not need to use Weierstrass's canonical factors. By the triangle inequality, 
$$
\left|~
\left(\frac{1}{2!}-\frac{1}{1!}\right)z^2
+\left(\frac{1}{3!}-\frac{1}{2!}\right)z^3
+\left(\frac{1}{4!}-\frac{1}{3!}\right)z^4
+\cdots\ \right|\\
\leqslant\frac12|z|^2+
\left|\left(\frac{1}{3!}-\frac{1}{2!}\right)z^3
+\left(\frac{1}{4!}-\frac{1}{3!}\right)z^4
+\cdots\right|
$$
Thus it suffices to show that
$$
\left|\left(\frac{1}{3!}-\frac{1}{2!}\right)z^3
+\left(\frac{1}{4!}-\frac{1}{3!}\right)z^4
+\cdots\right|\leqslant\frac12|z|^2
$$
Factoring out $z^3$ on the left hand side, applying the triangle inequality again, one has
$$
|z|^3\left[\left|\frac{1}{3!}-\frac{1}{2!}\right|
+\left|
\frac{1}{4!}-\frac{1}{3!}\right|
+\cdots\right]=|z|^3\cdot\frac{1}{2}
$$
Since $|z|<1$, we are done. 
One can justify the "$\cdots$" part by a limiting argument. 
</s>