Nilpotent ring and Nilpotent groups. 
Let $R$ be a ring (associative and with unity) and $B$ be a subring with the property that $B^n = 0$ i.e.
  $$ \forall\; x_1, x_2, \dots, x_n \in B: \; x_1 \cdot x_2 \cdots x_n = 0$$
  My aim is to prove that the set $G = \{1+x \mid x\in B\}$ is a nilpotent group.

It is easy to see that $G$ is a group. For $n=3$, I found
$$[1+x,1+y]_G = 1+[x,y]_R,$$ 
where $$[1+x,1+y]_G = (1-x+x^2)(1-y+y^2)(1+x)(1+y)$$ is the group commutator
and $[x,y]_R = xy-yx$ is the ring commutator. And $G$ is a nilpotent group of degree no more than $3$.
But if $n=4$, I got
$$[1+x, 1+y]_G = 1 + x^2y+xy+xyx+y^2x-yx-yxy,$$ and I don't know what to do. 
 A: First, note that since every element of $B$ is nilpotent, we have the following explicit formula for the inverse of $1-x$:
$$(1-x)^{-1}=1+x+x^2+...+x^{n-1}$$
So let's define our descending series of normal subgroups as:
$$G_k=1+B_k$$ Where $B_k$ is the additive subgroup of $B$ generated by elements of the form $x_1 x_2 ...x_m$ for $x_i\in B$, $m\geq k$.
One may check that $G_k$ is a normal subgroup in $G$, and we have $G=G_1$ and $G_n=1$.
So to show $G$ is nilpotent, it suffices to show that $[g,G_k]\subset G_{k+1}$ for all $g\in G$, where $[,]$ is the group commutator.
Doing this in the case of the element of $G_k$ being $1-x_1 x_2 ....x_m$:
$(1-y)(1-x_1 x_2 .... x_m)(1+y+y^2...+y^{n-1})(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$\big(1-(1-y)x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})\big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$=
$\big(1-x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})+y(x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})\big)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
$\alpha=-x_1 x_2 ..x_m(y+y^2+...y^{n-1})$
$\beta= y x_1 x_2 ...x_m(1+y+y^2+...y^{n-1})$
=$(1-x_1 x_2 ..x_m +\alpha +\beta)(1+x_1 x_2 ..x_m + (x_1 x_2...x_m)^2+...)$
=$1-(x_1...x_m)^2+(\alpha+\beta)\big((x_1 x_2...x_m)^2+...(x_1...x_m)^{n-1}\big)+(1-x_1 x_2 ..x_m)(x_1 x_2...x_m)^2+...)+(\alpha+\beta)(1+x_1 ...x_m)$
Which is in $G_{k+1}$, since all the terms after the $1$ in this expression are of length at least $m+1$.
The general element case is the exact same, its just a bit more of a mess.
