Having $f(z)=\sum\limits_{n=0}^{+\infty} \frac{1}{n!}z^n$
I had to find what $\sum\limits_{n=0}^{+\infty} \frac{1}{n!}z^n\sum\limits_{n=0}^{+\infty} \frac{D_n}{n!}z^n=\sum\limits_{n=0}^{+\infty} z^n$ was and it seems that it is
$$e^zf(z)=\frac{1}{1-z}$$
I understand the left size of the equation (even though I only feel that $\sum\limits_{n=0}^{+\infty} \frac{1}{n!}z^n=e^z$) but I don't understand the right one.
Thus why does $\sum\limits_{n=0}^{+\infty} z^n=\frac{1}{1-z}$
This is true only if $\lvert z\rvert <1$. The reason is quite simple and relies on a well known identity from high school: $$1-z^n=(1-z)(1+z+z^2+\dots+z^{n-1})$$ which we can rewrite as $$\frac1{1-z}=1+z+z^2+\dots+z^{n-1}+\frac{z^n}{1-z}.$$ From this we deduce that $$\biggl\lvert\frac1{1-z}-(1+z+z^2+\dots+z^{n-1})\biggr\rvert=\frac{\bigl\lvert z\rvert^n}{\lvert 1-z\rvert}\to 0\quad\text{as}\enspace n\to\infty.$$