$I-AB$ be invertible $\Leftrightarrow$ $I-BA$ is invertible

Hint:$(I-BA)^{-1}=X$ (say), Now expand left side. we get $$X=I+BA+ (BA)(BA)+(BA)(BA)(BA)+\dots$$ $$AXB=AB+(AB)^2+(AB)^3+(AB)^4+\dots$$ $$I+AXB=I+(AB)+(AB)^2+\dots+(AB)^n+\dots=(I-AB)^{-1}$$

Check yourself: $(I+AXB)(I-AB)=I$ and $(I-AB)(I+AXB)=I$

Hint :$(I - BA)(I + B(I - AB)^{-1}A) = I$

Claim: $\det(I-AB)=\det(I-BA)$.

Proof:

Case 1: $A$ is invertible. Exercise!

Case 2: $A$ is not invertible. Consider $k$ the algebraic closure of $F$.

Let $P(X)=\det(I-(A-xI)B)- \det(I-B(A-xI))$. By part Case 1, $P(x)=0$ for all $\{ x \in k| x \mbox{ is not an eigenvalue of} A \}$. Since $P(x)$ is a polynomial of degree $n$ which has infinitely many roots, it follows that $P(X) \equiv 0$, and hence $P(0)=0$.