Be $X$ a projective, geometrically connected and smooth curve over the finite field $\mathbb{F}_q$. Then one can define the following series, i.e. the zeta function of the curve: $$Z(t) = \exp\left(\sum_{d} \frac{\left|X(\mathbb{F}_{q^d})\right|}{d}t^d\right)$$ I would like to show that the following series (where, hopefully, there are no previous mistakes in its computation) is equal to the zeta: $$\mathcal{E}(t) = 1 + \sum_{d>0} q^d t^d \sum_{\sum_i n_i \deg(x_i)=d} \prod (1-q^{-\deg(x_i)})$$ where the inner sum is taken over all possible combinations of distinct closed points $x_i$ and positive integers $n_i$ with the property that $\sum_i n_i \deg(x_i) = d$, and the product is then taken over all the $x_i$ appearing in the sum.
I tried rewriting the series as the product over all $k$-tuple of points $x_i$, $k$ varying, of the sums of terms $1-q^{-\deg(x_i)}$, the sum being taken over the $k$-tuples of positive integers $(n_1, \dots, n_k)$, but i could not find an easier expression. In particular, the presence of the powers of $q$ throws me off, it seems like they should cancel out but i am not convinced how this should happen.
The question arose in the study of Hall algebras of curves, so if you have any idea how to use Hall-theoretic methods to simplify the problem, feel free to share your thought. I have tried some other other approaches using formal properties of the Hall algebras, like the Green product, but to no avail, and i believe there should be an "elementary" way to show the equivalence.
Thanks in advance for any comment or answer.