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? 
 A: There is a general phenomenon going on, which is encoded in Weierstrass's result about his elementary factors: 

if we set $E_n(z) = (1-z)e^{z+z^2/2+\cdots+z^n/n}$, then $|E_n(z)-1|\leqslant |z|^{n+1}$ when $|z|\leqslant 1$. 

To prove it, you can begin by noting that $$E_n'(z) =  -z^ne^{z+z^2/2+\cdots+z^n/n}$$ and that if $t\geqslant 0$, we have $|E_n'(tz)|\leqslant -|z|^n E_n'(t)$ because $|\exp z|\leqslant \exp |z|$. From this it follows now that
$$\begin{align}
|E_n(z)-1| &= \left|z \int_0^1 E_n'(zt)dt \right| \\
&\leqslant -|z|^{n+1} \int_0^1 E_n'(t)dt\\
&=-|z|^{n+1}(E_n(1)-E_n(0)) \\
&= |z|^{n+1}
\end{align}$$
There is a second proof obtained by looking at the Taylor expansion: again, because the Taylor expansion of $E_n'$ starts at $z^n$, the expansion of $E_n-1$ starts at $z^{n+1}$, say $E_n(z) = 1+ \sum_{k>n} a_k z^k$ (because $E_n(0)=1$). Second, all the coefficients $a_k$ are negative from the expression of $E_n'$ and real so $|a_k| = -a_k$, and $E_n(1) = 0$ implies that $- 1 = \sum_{k>n} a_k$. Thus if $|z|\leqslant 1$ we have
$$|E_n(z)-1| \leqslant \sum_{k>n} |a_k||z|^k \leqslant -|z|^{n+1}\sum_{k>n}a_k = |z|^{n+1}$$
A: 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. 
