Consider a function $f(x)$. Define Taylor series $\sum_{n=0}^{\infty} f(n) x^n$. Is there such a function, other than constant $0$, that $\sum_{n=0}^{\infty} f(n) x^n = f(x)$?

The Taylor series of $f(x)$ at $0$ is $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$. The series is unique, so $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^{\infty} f(n) x^n.$ This means that $f^{(n)}(0)=f(n) n!$. The constant function $0$ meets this condition. Are there others?

Add: There is an almost similar question here: Is there a function with the property $f(n)=f^{(n)}(0)$? It is not same, but looks a bit similar. Can it be used in any way?