Nilpotency class of the unitriangular matrix group I'm trying to compute the nilpotency class $c$ of unitriangular matrix group http://groupprops.subwiki.org/wiki/Unitriangular_matrix_group using the definition that $ [ ...[x_1, x_2 ] , x_3 ] ,..., x_{c + 1} ] = e $ for all $x_i \in G$.
I know the result is $ c =n -1 $ (from that page), and I think each step I should get a matrix with one more "layer" above the diagonal equal to $0$. But the computation seems very complicated. Is there a nice way to show this? 
Any help is appreciated. Many thanks in advance.
 A: You have to consider a formula in your link:
$$[e_{ij}(\lambda), e_{jk}(\mu)]=e_{ik}(\lambda\mu).$$
For simplicity, you may proceed with $n=3$, then $n=4$, and then in general.
Consider $n=4$, and $G=U(4,F)$. Then the generators of this group are 
$$e_{12}(\alpha), e_{13}(\alpha), e_{14}(\alpha), e_{23}(\alpha), e_{24}(\alpha), e_{34}(\alpha), \,\,\,\, \alpha\in F.$$
For nilpotency, it is sufficient to work on generators, and their commutators (in rough sense). By above formula, you can see that the commutators of the above generators are inside the normal subgroup
$$
\begin{Bmatrix}
\begin{bmatrix}
1 & 0 & * & *\\
  & 1 & 0 & *\\
  &   & 1 & 0 \\ 
 & & & 1 
\end{bmatrix}\colon * \in F.
\end{Bmatrix}.
$$
Conversely, every element of this subgroup is a member of commutator group, since consider its generators 
$$ (*) \,\,\,\,\,\, e_{13}(\alpha), e_{14}(\alpha), e_{24}(\alpha).$$
They can be expressed as commutators:
$$e_{13}(\alpha)=[e_{12}(\alpha), e_{23}(1)].$$
Try to express $e_{14}(\alpha), e_{24}(\alpha)$ as commutators. 
Thus, above subgroup is equal to the commutator subgroup $[G,G]$.
To talk about commutators of length $3$, i.e. of the form $[[x,y],z]$, again, consider $[x,y]$ from generators of $[G,G]$ (in (*)) and the generators of $G$ as taken initially. You will notice, from commutator formula that the "three-commutators" (i.e. $[[x,y],z]$) will generate the normal subgroup 
$$
\begin{Bmatrix}
\begin{bmatrix}
1 & 0 & 0 & *\\
  & 1 & 0 & 0\\
  &   & 1 & 0 \\ 
 & & & 1 
\end{bmatrix}\colon * \in F.
\end{Bmatrix}.
$$
This is sufficient to solve your problem.
