Proving $\operatorname{Tr}((AB)^m)=\operatorname{Tr}((BA)^m)$

Question

We know that $\operatorname{Tr}(AB)=\operatorname{Tr} (BA)$ How can we prove that $\operatorname{Tr}((AB)^m)=\operatorname{Tr}((BA)^m)$ ? I tried to use induction but it seemed that in the last step for $m+1$ I had to use the result and that was wrong. Can anyone explain a nicer way? Any hint can help. Thank you

Answer 1

Use $\operatorname{Tr}UV=\operatorname{Tr}VU$ with $U:=A,\,V:=(BA)^{m-1}B$.

Answer 2

Hint

Let's try for the case $Tr[(AB)^2]=Tr[(BA)^2]$.

$$Tr[(AB)^2]=Tr[(AB)(AB)]=Tr[ABAB]=Tr[A(BAB)]=$$ $$Tr[(BAB)A]=Tr[BABA]=Tr[(BA)^2]$$

Did you get it? Did you understand how to approach the general case?