I was trying to prove on my own this generalization of the more simple equality from matrix algebra $(AB)^{-1}=B^{-1}A^{-1}$, but it seems i got stuck when i was doing the induction, also i remember something about this problem beeing very easy to prove if you are allowed to use projectors or something like that, i think heard about it from a professor but i might be confusing this with some other thing, can someone help me whith this, and if possible tell me if this proof of using the projectors of the matrices makes any sense ?
$$ \left( \prod_{k=0}^{n}A_k \right)^{-1}= \prod_{k=0}^{n}A_{n-k}^{-1} $$
Clearly the result is true for $ n = 0, 1, 2 $. Suppose it's true for $ n $, and observe that $$ (\prod_{k = 0}^{n+1} A_k)^{-1} = (\prod_{k=0}^{n}A_k \cdot A_{n+1})^{-1}, $$ and using the the equality $ (AB)^{-1} = B^{-1}A^{-1} $ together with the induction hypothesis gives $$ (\prod_{k = 0}^{n+1} A_k)^{-1} = A_{n+1}^{-1} \cdot (\prod_{k=0}^{n} A_k)^{-1} = A_{n+1}^{-1} \cdot \prod_{k=0}^n A_{n-k}^{-1} = \prod_{k=0}^{n+1} A_{n-k}, $$ where the last equality follows from first shifting the index by $ 1 $ then combine the $ A_{n+1}^{-1} $ term.