A function satisfying a series equation We already solved or argued differential equations, integral equations and even function equations. But I encounter the following, which I call it "series equation":

Is there any nonzero function $f$ satisfying the series equation
  $$f(x)=\sum_{n=0}^{\infty}\frac{f(n)}{n!}x^n \tag{*} $$

The source: I interested to solve the equation $$f'(x)=f(x+1)$$
and I found out
$$f^{(n)}(x)=f(x+n)$$
thus: $f^{(n)}(0)=f(n)$ and if we consider  $f$ having a taylor series, then the expression $(*)$ will be the case.
But I have no idea how $f$ would be. Is such a function exists?
Thanks.
 A: Since nobody has answered the question yet, I will show how I obtained my examples (see my comments).  I will start with a more general delay differential equation (DDE)
$$f'(x)=af(x+b),$$
where $a,b\in\Bbb C$ are given constants s.t. $a,b\ne 0$.  We first find some $f:\Bbb C\to \Bbb C$ that works.  
We make an assumption that $f(x)=e^{\lambda x}$ for some complex number $\lambda$.  Plug this into the DDE, we get
$$\lambda e^{\lambda x}=ae^{\lambda (x+b)}=ae^{\lambda b}e^{\lambda x}.$$
Consequently $\lambda=ae^{\lambda b}$.  Therefore
$$(-\lambda b)e^{-\lambda b}=-ab.$$
This means $-\lambda b=W_k(-ab)$ for some integer $k$.  So we can define
$$\lambda_k=-\frac{W_k(-ab)}{b}$$
for $k\in\Bbb Z$, and let
$$f_k(x)=e^{\lambda_k x}.$$
Then any complex span $f$ of the functions $f_k$ for $k\in \Bbb Z$ satisfies the DDE. 
Now suppose that $a,b\in\Bbb R$.  If you want examples of $f:\Bbb R\to \Bbb C$, then you can take $f$ to be any complex span of the functions $f_k$ for $k\in \Bbb Z$.  
We still suppose that $a,b\in\Bbb R$.  If you want examples of $f:\Bbb R\to \Bbb R$, then you can take $f$ to be any real span of the functions $\Re f_k$ and $\Im f_k$ for $k\in\Bbb Z$.
Now the functions we obtain above are analytic.  Therefore they will also satisfy the series condition
$$f(x)=\sum_{n=0}^\infty\frac{f(nb)}{n!}(ax)^n.$$
For other solutions to $f'(x)=af(x+b)$, we may start with an arbitrary smooth function $h:I\to \Bbb C$ where $I=[0,b]$ if $b>0$, and $I=[b,0]$ if $b<0$.  We need extra conditions on $h(x)$ at the boundary points $x=0$ and $x=b$ to make the DDE work when $x=kb$ where $k$ are integers.  That is, $ah^{(n)}(b)=h^{(n+1)}(0)$  for all integers $n\ge 0$.  Then, define $f(x)=h(x)$ for $x\in I$.  Now we can use $f'(x)=af(x+b)$ to extend $f$ outside $I$.
