Help with Induction to show an Equality of Linear Operators I am trying to solve the following:
Let $X$ be a finite dimensional complex inner product space. Let $A,B \in \mathcal{L}(X)$, the space of linear mappings from $X$ to $X$. Use induction to show for any $n\geq1$, $$A^n-B^n=\sum_{m=0}^{n-1}A^m(A-B)B^{n-m-1}$$
For the base cases for $n=1,2$, this is straightforward to show, since for $n=1$ we have $A^0(A-B)B^0=A-B$, and for $n=2$ we have $A^0(A-B)B+A(A-B)B^0=AB-B^2+A^2-AB=A^2-B^2$.
Now we case assume the induction hypothesis: Suppose for $n=k$, it is true that $$A^k-B^k=\sum_{m=0}^{k-1}A^m(A-B)B^{k-m-1}$$. I attempted the following: Using the hypothesis, we can write $$A^k=\sum_{m=0}^{k-1}A^m(A-B)B^{k-m-1}+B^k$$ Thus $$A^{k+1}=\sum_{m=0}^{k-1}A^{m+1}(A-B)B^{k-m-1}+AB^k$$ Similarly for $B$, we have $$B^k=A^k-\sum_{m=0}^{k-1}A^m(A-B)B^{k-m-1}$$ Thus $$B^{k+1}=A^kB-\sum_{m=0}^{k-1}A^m(A-B)B^{k-m}$$ Combining, we have $$A^{k+1}-B^{k+1}=\sum_{m=0}^{k-1}A^{m+1}(A-B)B^{k-m-1}+AB^k-A^kB+\sum_{m=0}^{k-1}A^m(A-B)B^{k-m}$$ But I am unsure how to proceed. Any help is appreciated.

We know that $A^n-B^n=\sum_{m=0}^{n-1}A^m(A-B)B^{n-m-1}$ holds for $n = 1$.  Assume it holds for $n = k$, then for $n = k+1$, we have
\begin{align}
  & A^{(k+1)}-B^{(k+1)} = A^{(k+1)}-B^{(k+1)}+A^kB-A^kB \\
=\ & A^kB-B^{(k+1)} + A^{(k+1)}-A^kB = (A^k-B^k)B + A^k(A-B) \\
=\ & \left(\sum_{m=0}^{k-1}A^m(A-B)B^{k-m-1}\right)B+A^k(A-B) \\
=\ & \sum_{m=0}^{k-1}A^m(A-B)B^{(k+1)-m-1}+A^k(A-B)B^{(k+1)-k-1}\\
=\ & \sum_{m=0}^{(k+1)-1}A^m(A-B)B^{(k+1)-m-1}
\end{align}
So $A^n-B^n=\sum_{m=0}^{n-1}A^m(A-B)B^{n-m-1}$ holds for $n = k+1$, thus it holds for all positive integer $n$. 
