Let $G$ be a finite abelian group. When $\prod_{g\in G\setminus 1} (1-g)$ vanishes in (say, complex) group algebra of $G$?
It is easy to see that for cyclic group $G$ such product does not vanish, since $G$ may be embedded into a (complex) field.
Oh, it looks like it is obvious: for non-cyclic $G$ there is no exact complex representation of $G$, so for each character $\chi$ there exists $1\ne g\in G$ such that $\chi(1-g)=0$, so our product vanishes.
I think the answer is "if and only if the group $G$ is not cyclic". Why?
1) An element of $\mathbb C\left[G\right]$ is zero if and only if it acts as zero on each irreducible representation of $G$ (since $\mathbb C\left[G\right]$ is the direct sum of the endomorphism rings of the irreducible representations).
2) An element of $\mathbb C\left[G\right]$ acts on an irreducible representation of $G$ either as zero or as an automorphism (because each irreducible representation of $G$ is $1$-dimensional, since $G$ is abelian).
Hence, for a product of the form $\prod_{g\in S}\left(1-g\right)$ to be zero, where $S$ is some ordered list of elements of $G$, it is necessary and sufficient that for each irreducible representation of $G$, there exists some $g\in S$ such that $1-g$ acts as zero on the representation, i. e. that $g$ acts as identity on the representation. Applied to a list $S$ containing all elements of $G\setminus 1$ (maybe several times), this means that the product $\prod_{g\in S}\left(1-g\right)$ is zero if and only if no irreducible representation of $G$ is faithful. Easy manipulations with roots of unity show this to hold if and only if $G$ is not cyclic.
The complex group algebra of a finite abelian group is isomorphic, as a $\mathbb C$-algebra, to a direct product of copies of $\mathbb C$: it does not then have nilpotent elements.
