Why does $\sum\limits_{n=0}^{\infty}\frac{nz^{n-1}}{n!} = \sum\limits_{n=0}^{\infty}\frac{z^n}{n!}$ If $f(z) = \sum\limits_{n=0}^{\infty}\frac{z^n}{n!}$ is convergent for all $z$, why does
$f'(z) = \sum\limits_{n=0}^{\infty}\frac{nz^{n-1}}{n!} = \sum\limits_{n=0}^{\infty}\frac{z^n}{n!} = f(z)$ ?
Shouldn't this mean that $z=n$ ?

We have
$$\begin{align}
\sum_{n=0}^\infty\frac{nz^{n-1}}{n!}&=\sum_{n=1}^\infty\frac{nz^{n-1}}{n!}\\\\
&=\sum_{n=1}^\infty\frac{z^{n-1}}{(n-1)!}\\\\
&=\sum_{n=0}^\infty\frac{z^{n}}{n!}
\end{align}$$
And we are done!
