Is there a function with the property $f(n)=f^{(n)}(0)$? Is there a not identically zero, real-analytic function $f:\mathbb{R}\rightarrow\mathbb{R}$, which satisfies
$$f(n)=f^{(n)}(0),\quad n\in\mathbb{N} \text{ or } \mathbb N^+?$$
What I got so far:
Set
$$f(x)=\sum_{n=0}^\infty\frac{a_n}{n!}x^n,$$
then for $n=0$ this works anyway and else we have
$$a_n=f^{(n)}(0)=f(n)=\sum_{k=0}^\infty\frac{a_k}{k!}n^k.$$
Now $a_1=\sum_{k=0}^\infty\frac{a_k}{k!}1^k=a_0+a_1+\sum_{k=2}^\infty\frac{a_k}{k!},$ so 
$$\sum_{k=2}^\infty\frac{a_k}{k!}=-a_0.$$
For $n=2$ we find 
$$a_2=\sum_{k=0}^\infty\frac{a_k}{k!}2^k=a_0+a_1+2a_2+\sum_{k=3}^\infty\frac{a_k}{k!}2^k.$$
The first case was somehow special since $a_1$ cancelled out, but now I have to juggle around with more and more stuff. 
I could express $a_1$ in terms of the higher $a's$, and then for $n=3$ search for $a_2$ and so on. I didn't get far, however. Is there a closed expression? My plan was to argue somehow, that if I find such an algorythm to express $a$'s in terms of higher $a$'s, that then, in the limit, the series of remaining sum or sums would go to $0$ and I'd eventually find my function. 
Or maybe there is a better approach to such a problem.
 A: Let complex number $c$ be a solution of $e^c=c$.  For example $c = -W(-1)$, where $W$ is the Lambert W function.  Then since function $f$ defined by 
$$
f(x) = \sum_{n=0}^\infty \frac{e^{cn}x^n}{n!}
$$  
evaluates to $e^{e^c x} = e^{cx}$, we have $f(n) = e^{cn} = f^{(n)}(0)$.  For a real solution, let $c = a+bi$ be real and imaginary parts and let $g(x)$ be the real part of $f(x)$.   More explicitly:  
$$
g(x) = \sum_{n=0}^\infty \frac{e^{an}\cos(bn)\;x^n}{n!}
$$  
evaluates to $e^{ax}\cos(bx)$.  With the principal branch of Lambert W, this is approximately:  
$$
g(x) = e ^{0.3181315052 x} \operatorname{cos} (1.337235701 x)
$$
A: (not yet an answer, but too long for a comment) 
upps, I see there was a better answer of G.Edgar crossing. Possibly I'll delete this comment soon 
For me this looks like an eigenvalue-problem.
Let's use the following matrix and vector-notations. 
The coefficients of the power series of the sought function f(x) are in a columnvector A.
We denote a "vandermonde"-rowvector V(x) which contains the consecutive powers of x, such that $\small V(x) \cdot A = f(x) $
We denote the diagonal vector of consecutive factorials as F and its reciprocal as f
Then we denote the matrix which is a collection of all V(n) of consecutive n as ZV    .
Then we have first
$\qquad \small ZV \cdot A = F \cdot A  $
and rearranging the factorials
$\qquad \small  f \cdot ZV \cdot A = A  $
which is an eigenvalue-problem to eigenvalue $\small \lambda=1 $ . Thus we have the formal problem of solving
$\qquad \small  \left(f \cdot ZV - I \right) \cdot A = 0  $
However, at the moment I do not see how to move to the next step...

[added] The solution of G. Edgar gives the needed hint     

If we do not rearrange, but expand:      
$\qquad \small \begin{eqnarray}
   ZV \cdot A &=& F \cdot A  \\
   ZV \cdot (f \cdot F)\cdot A &=& F \cdot A  \\
   (ZV \cdot f) \cdot (F \cdot A) &=& (F \cdot A) \end{eqnarray} $     
we get a better ansatz. Let's denote simpler matrix constants $\small W=ZV \cdot f \qquad B=F \cdot A $ and rewrite this as
$\qquad \small W \cdot B = B $.
Now W is the carleman-matrix which maps $\small x \to \exp(x) $ by     
$\qquad \small W \cdot V(x) = V(\exp(x)) $    
Thus if $\small x = \exp(x) $ for some x , so x is some (complex) fixpoint $\small t_k$  of $\small f(x)=\exp(x)$ and we have one possible sought identity:      
$\qquad \small W \cdot V(t_k) = V(t_k) \to B = V(t_k) \to A = f \cdot B = f \cdot V(t_k) $  
Then the coefficients of the power series are      
$\qquad \small f(x) = 1 + t_k x + t_k^2/2! x^2 + t_k^3/3! x^3 + \ldots $
and $\small f(x) = \exp(t_k \ x) $
Because there are infinitely many such fixpoints (all are complex) we have infinitely many solutions of this type (there might be other types, the vectors $\small V(t_k) $ need not be the only type of possible eigenvectors of W )
A: I thought the problem was interesting and there seem to be a lot of guessing to come up with a solution. So I would just like to try to supplement with my own dubious and handwavy approach separate from GEdgar to motivate the solution, that might be more direct.
$$f(n) = f^{(n)}(0)$$
I then do a bit of arrangement by rewriting in terms of a translation operator,
$$\left.T^n f(x)\right|_{x=0} = \left.\frac{d^n}{dx^n} f(x)\right|_{x=0}$$
It is sometimes valid to write a translation operator as $T = e^{\frac{d}{dx}}$ and so,
$$\left.(e^{\frac{d}{dx}})^n f(x)\right|_{x=0} = \left.\left(\frac{d}{dx}\right)^n f(x)\right|_{x=0}$$
From here I consider the operator equation:
$$e^D = D$$
Which has the "solution" $D = -W(-1)$
Now I can use this to create an ansatz of the form $f(x) = Ke^{-W(-1)x}$ which we can then check in a fairly straightforward manner by simply plugging into the original 
$f(n) = f^{(n)}(0)$ and seeing that it does work out.
