Here's a fragment of something I posted in an answer a few months back:
\begin{align} & \left( 1 + \frac 1 {a_1} + \frac 1 {a_1^2} + \frac 1 {a_1^3} + \cdots \right) \\ \times {} & \left( 1 + \frac 1 {a_2} + \frac 1 {a_2^2} + \frac 1 {a_2^3} + \cdots \right) \\ \times {} & \left( 1 + \frac 1 {a_3} + \frac 1 {a_3^2} + \frac 1 {a_3^3} + \cdots \right) \\ \times {} & \quad \cdots \cdots \\ \vdots~ \end{align} When you expand the product, you multiply a term from the first factor, a term from the second factor, a term from the third factor, etc., but all except finitely many of those are $1$. The reason all but finitely many are $1$ is that if you multiply infinitely many non-$1$s, then the product is $0$, since $[\cdots]$
All this was intended to show the product is a sum whose every term is a product of finitely many of these terms.
Just supposing one could not assume these products were $0$, is there some neat formula for the expanded product?
The product $$\prod_{k=1}^N \Big( 1 + \frac{1}{a_k} + \frac{1}{a_k^2} + \cdots \Big)$$ expands formally to become the sum $$\sum_{\alpha} \frac{1}{a_{\alpha}}$$ over all multisets $\alpha$ containing $1$ through $N$. In other words it's the sum $$\sum_{e_1,...,e_N \in \mathbb{N}_0} \frac{1}{a_1^{e_1} \cdots a_N^{e_N}}.$$ It's easy to prove this by induction on $N$, since each multiset on $1,\ldots,N$ is a union in exactly one way of a multiset on $1,\ldots,N-1$ and a multiset on $N$.
As you let $N$ get larger you are just including more and more terms (but only ever finite monomials), so the series as $N \rightarrow \infty$ becomes the sum of reciprocals of all finite monomials.