Show that $ f(x)=c*e^x$ if $f'=f$ , but f is from a ring of power series over Complex Numbers, Context: generating Functions I am not sure what to do? I would simply say that i have a differential equation if the condition that f is from ring of power series. Its in the context of generating funktions.
 A: It appears I, having tackled this one on only my first sip of coffee for the day, got things slightly mixed up and solved a slight generalization of the requested result, that is, I showed that $f'(x) = cf(x)$ implies $f(x) = f(x_0) e^{cx}$.  Rather than go back and massively edit this answer, I hereby post it as is and go back to my coffee!  The requested answer should be evident from what follows.  Thank your for your patience.  Slurp! 
Without going into the rigamarole about why and when this series and the series of derivatives of the $a_i x^i$ properly converges, one calculates as follows:
$f(x) = \displaystyle \sum_0^\infty a_i x^i; \tag 1$
$f'(x) = \displaystyle \sum_1^\infty a_i i x^{i - 1}; \tag 2$
then if
$f'(x) = c f(x) \tag 3$
we obtain
$f'(x) = \displaystyle \sum_1^\infty a_i i x^{i - 1} = c \sum_0^\infty a_i x^i = \sum_0^\infty c a_i x^i; \tag 4$
comparing coefficients of like powers of $x$ we see that
$a_1 = ca_0, \tag 5$
$2a_2 = ca_1 = c^2 a_0 \Longrightarrow a_2 = \dfrac{1}{2}c^2 a_0, \tag 6$
$3a_3 = ca_2 \Longrightarrow a_3 = \dfrac{1}{2 \cdot 3}c^3 a_0, \tag 7$
$4a_4 = ca_3 \Longrightarrow a_4 = \dfrac{1}{2 \cdot 3 \cdot 4}c^4 a_0, \tag 7$
$5a_5 = ca_4 \Longrightarrow a_5 = \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5}c^5 a_0, \tag 8$
$\vdots \tag 9$
and in general, with $n \ge 1$,
$n a_n = ca_{n - 1} \Longrightarrow a_n = \dfrac{1}{2 \cdot 3 \cdot 4 \cdot \ldots \cdot n}c^n a_0 = \dfrac{1}{n!} c^n a_0; \tag{10}$
we may in fact deploy induction to prove (10), to wit:
$(n + 1)a_{n + 1} = ca_n \Longrightarrow a_{n + 1} = \dfrac{1}{n! \cdot (n + 1)} c^{n + 1} a_0 = \dfrac{1}{(n + 1)!} c^{n + 1} a_0; \tag{11}$
inserting (10) into (1) we find
$f(x) = \displaystyle \sum_0^\infty a_i x^i = a_0 + \sum_1^\infty \dfrac{1}{n!} c^n a_0 x^n = a_0 + a_0 \displaystyle \sum_1^\infty \dfrac{1}{n!} (cx)^n$
$= a_0 \left ( 1 + \displaystyle \sum_1^\infty \dfrac{1}{n!}(cx)^n \right ) = a_0 \displaystyle \sum_0^\infty \dfrac{1}{n!}(cx)^n = a_0 e^{cx}; \tag{12}$
the coefficient $a_0$, not being determined via (5)-(11), may be freely selected; if we take a differential equations point of view, it is determined by an as yet unsupplied initial condition, viz. $f(x_0)$; then
$f(x_0) = a_0 e^{c x_0}, \tag{13}$
or
$a_0 = f(x_0) e^{-cx_0}, \tag{14}$
whence
$f(x) = a_0 e^{cx} = f(x_0) e^{-c x_0} e^{c x} = f(x_0) e^{c(x - x_0)}; \tag{15}$ 
we often select $x_0 = 0$; then
$f(x) = f(x_0) e^{cx}. \tag{16}$
As for the precise question as asked, taking $c = 1$ in the above addresses that issue.  
Recursion before breakfast--meh!
A: If $f(x)=\sum_{n\geq 0} a_n x^n/n!$, then term by term differentiation implies that $f'(x)=\sum_{n\geq 0}a_{n+1} x^n/n!$. In particular if $f'=f$, then comparing coefficients of $x^n$ yields that
$$
a_n=a_{n+1} 
$$
for all $n\geq 0$ i.e. $a_n=a_0=f(0)$ for $n\ge 0$. Thus 
$$
f(x)=\sum_{n\geq 0}f(0)\frac{x^n}{n!}=f(0)e^x
$$ 
as desired.
