Let the following be a long exact sequence of free $A$-modules of finite rank:
$$0\to F_1\to F_2\to F_3\to...\to F_n\to0$$
I want to show that $\otimes_{i=1}^n (\det F_i)^{-1^{i}} \cong A$, where the notation $^{-1}$ means taking the dual.
My attempt was to break this into SES's like
$$0\to F_i\to F_i\oplus F_{i+1}\to F_{i+1}\to 0$$
from which we know that $$\frac {\det F_i\det F_{i+1}}{\det (F_i\oplus F_{i+1})}\cong A\tag{1}$$
Let me use the notation $$\begin{aligned}d_i &:= \det F_i\\ d_{i+(i+1)} &:= \det(F_i\oplus F_{i+1})\end{aligned}$$
From $(1)$, one readily gets that $d_i d_{i+1} \cdot d_{i+1} d_{i+2} = d_{i+(i+1)}d_{(i+1)+(1+2)}$ from which it follows that
$$\frac {d_i d_{i+2}}{d_{i+1}}=\frac{d_{i+(i+1)}d_{(i+1)+(1+2)}}{d_{i+1}^3}$$
But I am stuck at this step.
The determinant can be determined for every finitely generated projective module, because these are precisely the locally free modules of finite rank (which doesn't have to be constant, but it is locally constant, and on each constant piece we take the corresponding exterior power). It is additive on short exact sequences (see for example Daniel Murfet's notes). Now I claim that the statement holds more generally for locally free modules of finite rank (actually also for locally free sheaves of finite rank). This generalization is needed in order to make the induction work:
Let $K$ be the kernel of $F_{n-1} \to F_n$. We have an exact sequence $0 \to K \to F_{n-1} \to F_n \to 0$, which splits because $F_n$ is projective. Since $F_{n-1}$ is finitely generated projective, it follows that $K$ is finitely generated projective. Hence, $\det(K)^{-1} \otimes \det(F_{n-1}) \otimes \det(F_n)^{-1} \cong A$, i.e. $\det(K) \cong \det(F_{n-1}) \otimes \det(F_n)^{-1}$ We have the long exact sequence $$0 \to F_1 \to \dotsc \to F_{n-2} \to K \to 0.$$ By induction hypothesis, we have $$\det(F_1)^{-1} \otimes \dotsc \otimes \det(K)^{\pm 1} \cong A.$$ Now it follows $$\det(F_1)^{-1} \otimes \dotsc \otimes \det(F_{n-1})^{\pm 1} \otimes \det(F_n)^{\mp 1} \cong A.$$