Show that $f(z)=e^{f'(0)z}$ 
Let $f$ be holomorphic on a connected open set $U\subset\mathbb C$, we assume that $0\in U$ and for every $z,w\in U$ such that $z+w\in U$ one has $f(z+w)=f(z)f(w)$. Setting $z=0$ and considering the power series of $f$ around $w=0$, show that $f(z)=e^{f'(0)z}$ in a neighbourhood of $0$

I know that $f(0)=1$ and for every $z,w\in U$ such that $z+w\in U$: $f'(z+w)=f'(z)f(w)$
Now the power series of $f$ around $0$ is, $f(z)=\sum\limits_{n\ge0}\frac{f^{(n)}(0)}{n!}z^n$
to compare it with $e^{f'(0)z}$ I write the exponential as sum:
$e^{f'(0)z}=\sum\limits_{n\ge0}\frac{\left(f'(0)z\right)^n}{n!}$
so it must be valid that; $\left(f'(0)\right)^n=f^{(n)}(0)$
What does it mean, setting $z=0$ ?
 A: Hint: From $f(z+w)=f(z)f(w)$, you have
$$ f'(z+w)=f(z)f'(w). $$
Setting $w=0$ gives
$$ f'(z)=f'(0)f(z) \Longrightarrow \frac{f'(z)}{f(z)}=f'(0)$$
and then integrate both sides you will get the answer.
A: It is referring to your first statement after the question: having found that $f'(z+w)=f'(z)f(w)$, setting $z=0$ gives
$$ f'(w) = f'(0) f(w). $$
(Intuition: At this point the obvious thing to do is solve this differential equation, and we know it has solutions $Ae^{f'(0)w}$, so we must be able to deduce everything from here except the value of $f(0)$.)
Inductively, this tells us that
$$ f^{(n)}(w) = (f'(0))^n f(w). $$
(We just checked the $n=1$ case, and it should be clear how do do the $(n+1)$th case from the $n$th in the same way.)
Then setting $w=0$, you get $f^{(n)}(0) = (f'(0))^n f(0)$, which you can insert into the Taylor series formula to find
$$ f(w) = f(0) \sum_{n=0}^{\infty} \frac{(f'(0)w)^n}{n!} = f(0) e^{f'(0)w}. $$
Finally, finding $f(0)$ can be done by looking at $f(0+0)=f(0)f(0)$, so $f(0)=(f(0))^2$, and $f(0)=1$ or $0$. (It's not actually obvious here that we can exclude the $f \equiv 0$ case...)
