Problem Statement: Prove nilpotent group is supersolvable. A finite nilpotent group has supersolvable series
The link above uses $p$-group and direct product. But suppose we do not do that, the definition of niloptent seems to be only there EXISTS a central series. But can we claim that we can CONSTRUCT one as long as we follow $G/G_i \leq Z(G/G_{i-1})$?
I'm assuming your groups are finite. Let $1=G_0\le\ldots\le G_n=G$ be a central series. Since $G_i/G_{i-1}\le Z(G/G_{i-1})$, all subgroups of $G_i/G_{i-1}$ are normal in $G/G_{i-1}$. By the correspondence theorem, all subgroups of $G_i$ containing $G_{i-1}$ are normal in $G$. Since $G_i/G_{i-1}$ is abelian, we can refine our central series until $|G_i/G_{i-1}|$ is a prime for $i=1,\ldots,n$. But this means that $G$ is supersolvale.