Proof.
Consider the canonical map
\[ \pi ^*\mathcal{E} \longrightarrow \mathcal{O}_{\mathbf{P}(\mathcal{E})}(1) \]
and twist down by $1$ to get
\[ \pi ^*(\mathcal{E})(-1) \longrightarrow \mathcal{O}_{\mathbf{P}(\mathcal{E})} \]
This is a surjective map from a locally free rank $n + 1$ sheaf onto the structure sheaf. Hence the corresponding Koszul complex is exact (More on Algebra, Lemma 15.28.5). In other words there is an exact complex
\[ 0 \to \pi ^*(\wedge ^{n + 1}\mathcal{E})(-n - 1) \to \ldots \to \pi ^*(\wedge ^ i\mathcal{E})(-i) \to \ldots \to \pi ^*\mathcal{E}(-1) \to \mathcal{O}_{\mathbf{P}(\mathcal{E})} \to 0 \]
We will think of the term $\pi ^*(\wedge ^ i\mathcal{E})(-i)$ as being in degree $-i$. We are going to compute the higher direct images of this acyclic complex using the first spectral sequence of Derived Categories, Lemma 13.21.3. Namely, we see that there is a spectral sequence with terms
\[ E_1^{p, q} = R^ q\pi _*\left(\pi ^*(\wedge ^{-p}\mathcal{E})(p)\right) \]
converging to zero! By the projection formula (Cohomology, Lemma 20.54.2) we have
\[ E_1^{p, q} = \wedge ^{-p} \mathcal{E} \otimes _{\mathcal{O}_ S} R^ q\pi _*\left(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(p)\right). \]
Note that locally on $S$ the sheaf $\mathcal{E}$ is trivial, i.e., isomorphic to $\mathcal{O}_ S^{\oplus n + 1}$, hence locally on $S$ the morphism $\mathbf{P}(\mathcal{E}) \to S$ can be identified with $\mathbf{P}^ n_ S \to S$. Hence locally on $S$ we can use the result of Lemmas 30.8.1, 30.8.2, or 30.8.3. It follows that $E_1^{p, q} = 0$ unless $(p, q)$ is $(0, 0)$ or $(-n - 1, n)$. The nonzero terms are
\begin{align*} E_1^{0, 0} & = \pi _*\mathcal{O}_{\mathbf{P}(\mathcal{E})} = \mathcal{O}_ S \\ E_1^{-n - 1, n} & = R^ n\pi _*\left(\pi ^*(\wedge ^{n + 1}\mathcal{E})(-n - 1)\right) = \wedge ^{n + 1}\mathcal{E} \otimes _{\mathcal{O}_ S} R^ n\pi _*\left(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(-n - 1)\right) \end{align*}
Hence there can only be one nonzero differential in the spectral sequence namely the map $d_{n + 1}^{-n - 1, n} : E_{n + 1}^{-n - 1, n} \to E_{n + 1}^{0, 0}$ which has to be an isomorphism (because the spectral sequence converges to the $0$ sheaf). Thus $E_1^{p, q} = E_{n + 1}^{p, q}$ and we obtain a canonical isomorphism
\[ \wedge ^{n + 1}\mathcal{E} \otimes _{\mathcal{O}_ S} R^ n\pi _*\left(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(-n - 1)\right) = R^ n\pi _*\left(\pi ^*(\wedge ^{n + 1}\mathcal{E})(-n - 1)\right) \xrightarrow {d_{n + 1}^{-n - 1, n}} \mathcal{O}_ S \]
Since $\wedge ^{n + 1}\mathcal{E}$ is an invertible sheaf, this implies that $R^ n\pi _*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(-n - 1)$ is invertible as well and canonically isomorphic to the inverse of $\wedge ^{n + 1}\mathcal{E}$. In other words we have proved the case $d = - n - 1$ of the lemma.
Working locally on $S$ we see immediately from the computation of cohomology in Lemmas 30.8.1, 30.8.2, or 30.8.3 the statements on vanishing of the lemma. Moreover the result on $R^0\pi _*$ is clear as well, since there are canonical maps $\text{Sym}^ d(\mathcal{E}) \to \pi _* \mathcal{O}_{\mathbf{P}(\mathcal{E})}(d)$ for all $d$. It remains to show that the description of $R^ n\pi _*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(d)$ is correct for $d < -n - 1$. In order to do this we consider the map
\[ \pi ^*(\text{Sym}^{-d - n - 1}(\mathcal{E})) \otimes _{\mathcal{O}_{\mathbf{P}(\mathcal{E})}} \mathcal{O}_{\mathbf{P}(\mathcal{E})}(d) \longrightarrow \mathcal{O}_{\mathbf{P}(\mathcal{E})}(-n - 1) \]
Applying $R^ n\pi _*$ and the projection formula (see above) we get a map
\[ \text{Sym}^{-d - n - 1}(\mathcal{E}) \otimes _{\mathcal{O}_ S} R^ n\pi _*(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(d)) \longrightarrow R^ n\pi _*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(-n - 1) = (\wedge ^{n + 1}\mathcal{E})^{\otimes -1} \]
(the last equality we have shown above). Again by the local calculations of Lemmas 30.8.1, 30.8.2, or 30.8.3 it follows that this map induces a perfect pairing between $R^ n\pi _*(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(d))$ and $\text{Sym}^{-d - n - 1}(\mathcal{E}) \otimes \wedge ^{n + 1}(\mathcal{E})$ as desired.
$\square$
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