Lower central series of the Unitriangular group $UT(n, \mathbb{Z}_p)$ This is Exercise 5.44 from Rotman's book "An Introduction to the theory of Groups (4th Ed)". Specificaly, the exercise asks us to prove that the $i$-eth term in the lower central series is the set of all unitriangular matrices with "$0$" in the $i$ super-diagonals. In particular, this means the group is nilpotent of class $n-1$.
While I believe I could do it in a rather straightforward way by calculating the commutators explicitly (by hand), Rotman provides a hint: "Given $A \in UT(n, \mathbb{Z}_p)$, consider the powers of the matrix $A -I_n$".

What I don't get is how to use this hint...

I get that $A - I_n$ is nilpotent of class $n - z$, where "$z$" denotes the number of super-diagonals with only $0$. So, for instance, if $n = 2$ and $$A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$$
then $$A - I_2 = \begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix}$$
which is nilpotent of class $2$ ($(A-I_2)^2 = 0$). However,

how can this be used to solve the problem at hand, in a way to (hopefully) avoid too much explicit calculation?

Please provide only a hint (if possible).
Thanks in advance!
 A: I think this is what Rotman had in mind, to take commutators of matrices $A-E$ rather than powers. Let $A_{i,j}(a)$ be the matrix with a single entry $a$ in the $i,j$th position ($i<j$). The inverse of $I+A_{i,j}(a)$ is $I-A_{i,j}(a)$. This is easy once one notices that $A_{i,j}(a)^2=0$. We will need this later!
Simply compute the commutator of two of these:
$$[I+A_{i,j}(a),I+A_{k,l}(b)]=(I-A_{i,j}(a))(I-A_{k,l}(b))(I+A_{i,j}(a))(I+A_{k,l}(b)).$$
The product of the first two terms is $(I+A_{i,j}(a)A_{k,l}(b))-(A_{i,j}(a)+A_{k,l}(b))$. We won't worry about that product in there yet. But now we can write the whole thing as an $(x-y)(x+y)$ product to get simply $x^2-y^2$ for the quadruple product: we obtain
$$(I+A_{i,j}(a)A_{k,l}(b))^2-(A_{i,j}(a)+A_{k,l}(b))^2$$
Now we worry about the product. The square of $A_{i,j}(a)$ and $A_{k,l}(b)$ is zero. The product $A_{i,j}(a)A_{k,l}(b)$ is $0$ unless $j=k$, in which case it's $A_{i,l}(a+b)$, and its square is also zero. So now the first bracket becomes
$$I+2\delta_{j,k}A_{i,l}(a+b),$$
where $\delta_{i,j}$ is the standard delta-function. The second bracket's square terms are zero, and we obtain from the cross terms
$$ \delta_{j,k}A_{i,l}(a+b)+\delta_{i,l}A_{j,k}(a+b).$$
The first of these cancels with one of the two from the other term, and we obtain
$$[I+A_{i,j}(a),I+A_{k,l}(b)]=I+\delta_{j,k}A_{i,l}(a+b)-\delta_{i,l}A_{j,k}(a+b).$$
So from this we can easily determine a bunch of elements. In particular, for any $j>1$,
$$ [I+A_{i,i+1}(1),I+A_{i+1,i+j}(a)]=I+A_{i,i+j}(a).$$
This gets you a generating set for the set you want, given as a commutator of $UT$ and the previous term in the central series.
Writing $A=I+X$ for $X$ a nilpotent matrix helps a bit because squares are zero. But I'm not convinced there is an elementary, yet non-computational, way to do this. Someone might be along in a while to prove me wrong, of course.
