The Alternating Product of Determinant Bundles is Trivial

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This is Vakil 13.5 H, self-study.

We are to show that given an exact sequence of finite rank locally free sheaves on $X$

$$0 \to \mathcal F_1 \to \mathcal F_2 \to ... \to \mathcal F_n \to 0$$

we must have

$$\det \mathcal F_1 \otimes \det (\mathcal F_2)^{\vee} \otimes \det \mathcal F_3 \ \otimes ... \otimes \det \mathcal (F_n)^{(-1)^n} \simeq \mathcal O_X$$

where $\det \mathcal F := \bigwedge^{\text{rank} \ \mathcal F} \mathcal F$. The hint is to break up into short exact sequences and show those are short exact sequences of finite rank locally free sheaves. I have done so using a previous exercise, yielding short exact sequences of finite rank locally free sheaves

$$0 \to \operatorname{coker} \phi_{i - 2} \to \mathcal F_i \to \operatorname{image} \phi_i \to 0$$

where the $\phi_i : \mathcal F_i \to \mathcal F_{i + 1}$ come from the original exact sequence. We are then supposed to use the fact that for a short exact sequence $0 \to \mathcal F' \to \mathcal F \to \mathcal F'' \to 0$ of locally free finite rank sheaves,

$$\det \mathcal F \simeq \det \mathcal F' \otimes \det \mathcal F''$$

Using this hint, we obtain

$$\det \mathcal F_i \simeq \det (\operatorname{image} \phi_{i}) \otimes \det (\operatorname{coker} \phi_{i - 2})$$

Main question: how do I use this fact to conclude? I do not know how to envision the dual of the determinant, nor do I see how using the fact above gives us what we want.

Small side question: what does the $(-1)^n$ mean in the original exact sequence? I gather it is supposed to mean that we are alternating between dual and no dual, but I do not understand why that is appropriate notation.