Commutator of a matrix to the power of k The question asks me to show that $$[A,B^k] = \Sigma_{r=1}^k B^{r-1}[A,B]B^{k-r}$$ (A, B are nxn matrices) but I can't get even close. I suspect there's some definition of $B^k$ that I don't know but is required. I've tried expanding the RHS, to get
$$\Sigma_{r=1}^k B^{r-1}(AB-BA)B^{k-r}$$
$$= \Sigma_{r=1}^k B^{r-1}ABB^{k-r} - B^{r-1}BAB^{k-r}$$
So starting from the LHS, what can I do to $B^k$? Substituting in the diagonalised matrix such that $B^k = P\Lambda^k P^{-1}$ didn't get me anywhere and I'm not sure where the sum comes into it. The only sums I've seen in matrix calculations come from exp(B), but I don't think that's related. Any help or hints are much appreciated!

You are close.
\begin{align}
\sum_{r=1}^k B^{r-1}(AB-BA)B^{k-r}&=\sum_{r=1}^k (B^{r-1}AB^{k-r+1} - B^{r}AB^{k-r})
\\
&=\sum_{r=1}^k B^{r-1}AB^{k-r+1}-\sum_{r=1}^k B^{r}AB^{k-r}
\\
&=\sum_{r=0}^{k-1} B^{r}AB^{k-r}-\sum_{r=1}^k B^{r}AB^{k-r}
\\
&=AB^{k}+\sum_{r=1}^{k-1} B^{r}AB^{k-r}-\sum_{r=1}^{k-1} B^{r}AB^{k-r}-B^kA
\\
&=AB^{k}-B^kA
\\
&=[A,B^k]
\end{align}
As @Andreas comment, this holds not only for matrix, but for general ring. 
