Express the $(A + B)^k$ for $A, B \in \ M_{n \times n}(\mathbb{R})$ and $AB - BA = A$ Let $A, B \in \ M_{n \times n}(\mathbb{R})$
We know that $AB - BA = A$
How to express in terms of $A^iB^j$ the expression $(A + B)^k$
 A: Consider following two matrix valued functions
$\begin{cases}
U(t) &= e^{At}Be^{-At}\\
V(t) &= e^{Bt}Ae^{-Bt}
\end{cases}$.
Notice
$$U'(t) = e^{At}(AB - BA)e^{-At} = e^{At}Ae^{-At} = A\quad\implies\quad U(t) = B + At$$
We have
$$A + B = U(1) = e^A B e^{-A}\quad\implies\quad (A+B)^k = e^A B^k e^{-A}$$
Similarly,
$$V'(t) = e^{Bt}(BA-AB)e^{-Bt} = -e^{Bt}A^{-Bt} = -V(t)\quad\implies\quad
V(t) = e^{-t} A$$
From this, we can deduce
$$e^{Bt} e^{-A} e^{-Bt} = e^{-e^{Bt}Ae^{-Bt}} = e^{-e^{-t}A}
\quad\implies\quad e^{A} e^{Bt} e^{-A} = e^{(1-e^{-t})A} e^{Bt}$$
On the RHS, the horrible looking exponent in $A$ is very similar to the
generating function for the Touchard polynomials.
$$ e^{x(e^t-1)} = \sum_{n=0}^\infty \frac{T_n(x)}{n!} t^n\quad\text{ where }\quad
T_n(x) = \sum_{\ell=0}^n \left\{ {n \atop \ell }\right\} x^\ell$$
and $\left\{ {n \atop \ell }\right\}$ is the Stirling's number of second kind. From this, we find
$$e^A \left( \sum_{k=0}^\infty \frac{B^k}{k!}t^k\right) e^{-A} = 
\left(\sum_{n=0}^\infty \frac{(-1)^n T_n(-A)}{n!} t^n\right)
\left(\sum_{m=0}^\infty \frac{B^m}{m!} t^m\right)$$
By comparing the coefficient of $t^k$ on both sides, we get
$$(A+B)^k = e^A B^k e^{-A} = \sum_{n=0}^k (-1)^n \binom{k}{n} T_n(-A) B^{k-n}$$
A: When $[A,B]$ and $A$ commute, $e^ABe^{-A}=B+[A,B]$. Moreover,  $[A,B]=A$ implies that $A$ is nilpotent. Then $(A+B)^k=e^AB^ke^{-A}=\sum_{p=0}^{n-1}A^p/p!B^k\sum_{q=0}^{n-1}(-1)^qA^q/q!$
$=\sum_{p,q<n}\dfrac{(-1)^q}{p!q!}A^pB^kA^q$. It remains to calculate $B^kA^q$ as function of the $(A^iB^j)_{i,j}$. We obtain, by induction, $BA^p=A^pB-pA^p$ and more generally $B^kA^p=A^pB^k-\binom{k}{1}pA^pB^{k-1}+\binom{k}{2}p^2A^pB^{k-2}+\cdots+(-1)^kp^kA^p$. 
Note that if $p+q\geq n$, then $A^pB^kA^q=0$.
