Rational points on symmetric products and effective zero cycles Let $X$ be a variety defined over the finite field $\mathbb{F}_p$. The following is "well known". Let $a_n$ be the number of $\mathbb{F}_{p^{n}}$-rational points of $X$, and let $b_n$ be the number of $\mathbb{F}_p$-rational points of the $n$-fold symmetric product of $X$, i.e $X^n/S_n$ where $S_n$ is the symmetric group on $n$ letters. Then the following equality of rational power series holds:
$\sum{b_n t^n} = \exp(\sum{\frac{a_n}{n} \cdot t^n})$
The explanation is that there is a correspondence between $\mathbb{F}_p$ rational points on $X^n / S_n$ and degree $n$ effective zero cycles on $X$, and these in turn are related to rational points on $X$ in finite field extensions of $\mathbb{F}_p$.
I am not familiar with intersection theory and with algebraic cycles beyond the definitions in the appendix of Hartshorne, but I suppose that the case of degree zero should be easy to explain. I will be happy with an explanation in the case where $X$ is affine

You need to know nothing about intersection theory and very little about algebraic geometry to understand this result; it's almost entirely combinatorial. The combinatorial heart of this result is known as the exponential formula; see this blog post and this blog post for some details. 
"Degree $n$ effective zero cycle on $X$" just means a formal sum $\sum n_p p$ of points over $\overline{\mathbb{F}_p}$ where $n_p \ge 0$ are non-negative integers summing to $n$ and the sum is closed under Galois action. Such a thing has a unique "cycle decomposition" into Galois orbits, and that's all the algebraic geometry you need to know; the rest is the exponential formula.
