Question: 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.

Answer: 
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: 

$$
g(x) = e ^{0.3181315052 x} \operatorname{cos} (1.337235701 x)
$$
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.