B Elegant way to prove ##AB^{n}## commute

[askmathquestions]

If $A$ and $B$ are two matrices which commute, then probably $AB^{n} = B^{n}A$ too.

Now, I could probably do all the work of proving this with induction, but I feel like there should be a more elegant way to prove this, though I don't know exactly how to avoid writing ellipses. If I start with $AB^{n} = A \prod_{k=1}^{n}B = AB...B = BAB... = BBAB...B -> = BB...BA ...$ and we get this sort of iterative commutation argument, though I'm not sure how to formalize it.

 

[PeroK]

Induction is elegant, especially in cases like this!