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

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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.
 
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Induction is elegant, especially in cases like this!
 

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