I don't understand the italicized part in the proof of (ii):
Proposition Let $E$ be a ring (with $1$) and $I$ an ideal of $E$ with $I^n=0$. Put $$G=1+I=\{1+x, x\in I\} \subset E.$$ (i) $G$ is a subgroup of the group of units of $E$. (ii) Put $G_i=1+I^i$ for $i=1, ..., n$; then $$1=G_n\leq G_{n-1} \leq...\leq G_1=G$$ is a central series of $G$, so $G$ is nilpotent of class at most $n-1$.
Proof (i) For $x\in I$ we have $(1+x)(1-x+x^2-...\pm x^{n-1})=1$ so each element of $G$ is a unit in $E$; and $$(1+x)(1+y)=1+(x+y+xy)$$ so $G$ is closed under multiplication. (ii) Note that each $G_i$ is a subgroup of $G$, by the argument of part (i)...
I see that we would replace $x$ by $x^i$ in (i), but for instance, how do you know that $-x^i+x^{2i}-...\pm x^{i(n-1)}$ is in $I^i$? To show closure, how would you know that $x^i+y^i+x^iy^i$ is in $I^i$? So far it doesn't make sense to me.
This is proposition 9, page 7, in Daniel Segal's Polycyclic Groups.
