Can someone help me prove this please?
Suppose that $A_1,A_2,....A_m$ are all invertible $n \times n$ matrices. Then, $(A_1\cdot A_2,....\cdot A_m)^{-1}=A_1^{-1}\cdot.....\cdot A_{m-1}^{-1}\cdot A_{m}^{-1}$.
Base case $m=1$ : this is a tautology and true by inspection
Assume for some $k \in \mathbb N$ $(A_1 \cdot A_2,.... \cdot A_k)^{-1}=A_1^{-1}\cdot.....\cdot A_{k-1}^{-1}\cdot A_{k}^{-1}$.
Inductive step - let $m=k+1$: $(A_1 \cdot A_2,....\cdot A_k \cdot A_{k+1})^{-1} = ((A_1 \cdot A_2,....\cdot A_k)\cdot A_{k+1})^{-1} = A_{k+1}^{-1} \cdot (A_1 \cdot A_2,.... \cdot A_k)^{-1}$ by identity $(A\cdot B)^{-1} = B^{-1}\cdot A^{-1}$. Then I think the steps are obvious what I would do - replace $A^{-1}$ with the assumption and then use the commutativity of matrix multiplication.
Hence true by induction.
I think my inductive step is incorrect. Can I have some help please ?
For the base case, i.e. $m=2$, note that since $A_1$ and $A_2$ are invertible, their product is invertible and hence $(A_1 A_2)^{-1}A_1A_2 = I$. Multiplying both sides from the right by $A_2^{-1}$ and $A_1^{-1}$ respectively shows that $(A_1A_2)^{-1} = A_2^{-1}A_1^{-1}$. Now assume that $(A_1A_2\cdots A_k)^{-1} = A_k^{-1}A_{k-1}^{-1}\cdots A_1$ for some $k \in \mathbb{N}$. Then \begin{align} (A_1A_2\cdots A_kA_{k+1})^{-1} &= ((A_1 A_2\cdots A_k) A_{k+1})^{-1} \\ & = A_{k+1}^{-1} (A_1A_2 \cdots A_k)^{-1} \\ & = A_{k+1}^{-1} A_k^{-1} A_{k-1}^{-1} \cdots A_1^{-1}, \end{align} where we used associativity of matrix multiplication in the first step, the $m=2$ case in the second step and the induction hypothesis in the final step.