Alternative proof for the power of square commuting matrices

[diogomgf]

Besides mathematical induction, how would I approach the following problem:

Let \(\displaystyle A, B \in M_{n*n} \) such that \(\displaystyle A \cdot B = B \cdot A \).
Prove that: \(\displaystyle A \cdot B^k = B^k \cdot A\) for any \(\displaystyle k \in N_0 \) ?

 

[pka]

Besides mathematical induction, how would I approach the following problem:
Let \(\displaystyle A, B \in M_{n*n} \) such that \(\displaystyle A \cdot B = B \cdot A \).
Prove that: \(\displaystyle A \cdot B^k = B^k \cdot A\) for any \(\displaystyle k \in N_0 \) ?

To diogomgf, Why an alternative proof? Here are some hints that may be useful.
Given that \(\displaystyle A\cdot B=B\cdot A\) then if \(\displaystyle K>1\)
\(\displaystyle \begin{align*}A\cdot B^K&=A\cdot B\cdot B^{K-1} \\&=B\cdot A\cdot B^{K-1} \\&=B\cdot A\cdot B\cdot B^{K-2}\\&=B\cdot B\cdot A\cdot B^{K-2}\\&=B^2\cdot A\cdot B^{K-2} \end{align*}\)
What if \(\displaystyle K=2~?\)
I have no idea if that helps at all.

 

[diogomgf]

To diogomgf, Why an alternative proof? Here are some hints that may be useful.

Curiosity, I guess...
I was looking for something using the definition of matrix multiplication, if such proof is possible to muster...