<s>[INST] How to find power series of $f(z)=\frac{e^z}{1-z}$ at $z_0=0$? I tried to calculate few derivatives, but I cant get $f^{(n)}(z)$ from them. Any other way?   
$$f(z)=\frac{e^z}{1-z}\text{ at }z_0=0$$
 [/INST] $$
g(z) = a_0+a_1z+a_2z^2+...,\\
\frac{g(z)}{1-z}=a_0\frac{1}{1-z}+a_1\frac{z}{1-z}+a_2\frac{z^2}{1-z}+...
$$
Using the power series for $\frac{1}{1-z}$ gives
$$
g(z)=a_0(1+z+z^2+...)+a_1z\cdot(1+z+z^2+...)+a_2z^2\cdot(1+z+z^2+...),\\
g(z)=a_0(1+z+z^2+...)+a_1\cdot(z+z^2+z^3+...)+a_2\cdot(z^2+z^3+z^4...),\\
g(z)=a_0+(a_0+a_1)z+(a_0+a_1+a_2)z^2+...
$$
For the specific case $g(z)=e^z$
$$
\frac{e^z}{1-z}=\frac{1}{0!}+\left(\frac{1}{0!}+\frac{1}{1!}\right)z+\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}\right)z^2+...+\left(\sum_{r=0}^n\frac{1}{r!}\right)z^n+...
$$
as required.
</s>