<s>[INST] Power series and computation Let $F\colon \mathbb{C}\to\mathbb{C}$ and let $f\colon \mathbb{R}\to\mathbb{R}$ be the functions defined by: 
$$ F(z) = \sum_{n=0}^{\infty} \frac{z^n}{(n+1)!}$$
$$ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!}$$
Show that 
$$ F(z) = 
\begin{cases}
\frac{e^z-1}{z}  & \text{for $z\in\mathbb{C}\setminus \{0\}$} \\
1 & \text{if $z=0$}
\end{cases}$$
and show that $xf'(x) + f(x) = e^x$

It is trivial that for $F(z) = 0\ \text{if $z=0$}$
However, I am not sure how to show that for all $z\in\mathbb{C}$ $F(z) = \frac{e^z-1}{z}$. 
In order to show that $xf'(x) + f(x) = e^x$ I know that $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$ and $xf'(x) = \sum_{n=0}^{\infty}n\cdot \frac{x^{n-1}\cdot x}{(n+1)!}=\sum_{n=0}^{\infty}n\cdot \frac{x^{n}}{(n+1)!}$
$x'f(x) + f(x) = \sum_{n=0}^{\infty}n\cdot \frac{x^{n}}{(n+1)!}+\sum_{n=0}^{\infty} \frac{x^{n}}{(n+1)!} = \sum_{n=0}^{\infty}n\cdot \frac{x^{n}}{(n+1)!}$
Where do I go from here to get $e^x$?
 [/INST] $$F(z) = \sum_{n=0}^{\infty} \frac{z^n}{(n+1)!}=\frac{1}{z}\sum_{n=0}^{\infty} \frac{z^{n+1}}{(n+1)!}=\frac{1}{z}\sum_{\color{blue}{n=1}}^{\infty} \frac{z^{n}}{n!}=\frac{1}{z}(\color{blue}{e^z-1})$$
edit: For the second part,

$$xf'(x) + f(x)=x\sum_{n=0}^{\infty} \frac{nx^{n-1}}{(n+1)!}+\sum_{n=0}^{\infty} \frac{x^n}{(n+1)!}=\sum_{n=0}^{\infty} \frac{(n+1)x^{n}}{(n+1)!}=e^x$$ by cancelling $(n+1)$ from the fraction.

</s>