For a sequence of real numbers $\left(a_k\right)_{k \in \mathbb{N}}$, does the fact that
$$ g(x) = \sum_{k=0}^{\infty} a_k \cdot \frac{x^k}{k!} = 0, \forall x \in \{1, 2, 3, ... \} $$
imply that $a_k = 0, \forall k \in \{0, 1, 2, 3, ... \}$?
In the case where the property is true $\forall x \in \left(0, \infty \right)$, one can use differentiation and Taylor expansion (as was done in this answer) to conclude that $a_k = 0, \forall k = \{0, 1, 2, ... \}$.
I don't think the same approach works in the case where the values taken by $x$ are discrete.
I think it's enough to show that the infinite matrix
\begin{pmatrix} 1^0 & 1^1 & 1^2 & ... \\ 2^0 & 2^1 & 2^2 & ... \\ 3^0 & 3^1 & 3^2 & ... \\ . & . & . \\ . & . & . \\ . & . & . \\ \end{pmatrix}
has a determinant of $0$, but I'm not sure where to start. I don't think an approach related to generating functions will work either, since only the values at positive integer $x$ are known.
Can anyone give me a hint on how to proceed with this?
No, consider $$ \sin (\pi x) = \sum\limits_{k = 1}^\infty {( - 1)^{k - 1} \pi ^{2k - 1} \frac{{x^{2k - 1} }}{{(2k - 1)!}}} . $$ In this example, with your notation, $$ a_{2k - 1} = ( - 1)^{k - 1} \pi ^{2k - 1} \neq 0,\quad a_{2k} = 0, $$ yet the power series vanishes at integer values of $x$.