Let $g : \mathbb{N}_0 \rightarrow \mathbb{R}$, such that
$0 = \sum_{i \in \mathbb{N}_0} \frac{g(i)}{i!} \lambda^i$ for $\lambda > 0$.
A professor of mine said, that $g(i) = 0$ follows from the identity theorem for power series. However, in the theorem it is clearly stated that the power series has to be convergent in some neighbourhood of the center point of the series. In this case, the center point appears to be $0$, so the conditions for the identity theorem are not met. I have tried using a different center point on the negative half lines, but i keep ending up with negative values for $\lambda$ for any neighborhood of a center point.
I.e. $\lambda := z - a>0$ for some $a<0$, then $0 = \sum_{i \in \mathbb{N}_0} \frac{g(i)}{i!} (z-a)^i$ and the power series is supposed to be convergent for any $z \in B(a,\varepsilon)$, but $a-\varepsilon < z < a$ yields $\lambda < 0$...
If $g(i)\neq0$ for some $i\in\mathbb N_0$, let $j$ be the smallest element of $\mathbb N_0$ such that $g(j)\neq0$. Then\begin{align}0&=\frac0{\lambda^j}\\&=\frac{g(j)}{j!}+\frac{g(j+1)}{(j+1)!}\lambda+\cdots\\&=\frac{g(j)}{j!}\left(1+\frac{g(j+1)}{g(j)(j+1)}\lambda+\frac{g(j+2)}{g(j)(j+1)(j+2)}\lambda^2+\cdots\right)\end{align}when $\lambda\neq0$. But this is impossible, since$$\lim_{\lambda\to0}\frac{g(j)}{j!}\left(1+\frac{g(j+1)}{g(j)(j+1)}\lambda+\frac{g(j+2)}{g(j)(j+1)(j+2)}\lambda^2+\cdots\right)=\frac{g(j)}{j!}\neq0.$$