We will use the standard affine open covering $$ \mathcal{U} : \mathbf{P}^n_R = \bigcup\nolimits_{i = 0}^n D_{+}(T_i) $$ to compute the cohomology using the {\v C}ech complex. This is permissible by Lemma \ref{lemma-cech-cohomology-quasi-coherent} since any intersection of finitely many affine $D_{+}(T_i)$ is also a standard affine open (see Constructions, Section \ref{constructions-section-proj}). In fact, we can use the alternating or ordered {\v C}ech complex according to Cohomology, Lemmas \ref{cohomology-lemma-ordered-alternating} and \ref{cohomology-lemma-alternating-usual}. \medskip\noindent The ordering we will use on $\{0, \ldots, n\}$ is the usual one. Hence the complex we are looking at has terms $$ \check{\mathcal{C}}_{ord}^p(\mathcal{U}, \mathcal{O}_{\mathbf{P}_R}(d)) = \bigoplus\nolimits_{i_0 < \ldots < i_p} (R[T_0, \ldots, T_n, \frac{1}{T_{i_0} \ldots T_{i_p}}])_d $$ Moreover, the maps are given by the usual formula $$ d(s)_{i_0 \ldots i_{p + 1}} = \sum\nolimits_{j = 0}^{p + 1} (-1)^j s_{i_0 \ldots \hat i_j \ldots i_{p + 1}} $$ see Cohomology, Section \ref{cohomology-section-alternating-cech}. Note that each term of this complex has a natural $\mathbf{Z}^{n + 1}$-grading. Namely, we get this by declaring a monomial $T_0^{e_0} \ldots T_n^{e_n}$ to be homogeneous with weight $(e_0, \ldots, e_n) \in \mathbf{Z}^{n + 1}$. It is clear that the differential given above respects the grading. In a formula we have $$ \check{\mathcal{C}}_{ord}^\bullet(\mathcal{U}, \mathcal{O}_{\mathbf{P}_R}(d)) = \bigoplus\nolimits_{\vec{e} \in \mathbf{Z}^{n + 1}} \check{\mathcal{C}}^\bullet(\vec{e}) $$ where not all summands on the right hand side occur (see below). Hence in order to compute the cohomology modules of the complex it suffices to compute the cohomology of the graded pieces and take the direct sum at the end. \medskip\noindent Fix $\vec{e} = (e_0, \ldots, e_n) \in \mathbf{Z}^{n + 1}$. In order for this weight to occur in the complex above we need to assume $e_0 + \ldots + e_n = d$ (if not then it occurs for a different twist of the structure sheaf of course). Assuming this, set $$ NEG(\vec{e}) = \{i \in \{0, \ldots, n\} \mid e_i < 0\}. $$ With this notation the weight $\vec{e}$ summand $\check{\mathcal{C}}^\bullet(\vec{e})$ of the {\v C}ech complex above has the following terms $$ \check{\mathcal{C}}^p(\vec{e}) = \bigoplus\nolimits_{i_0 < \ldots < i_p, \ NEG(\vec{e}) \subset \{i_0, \ldots, i_p\}} R \cdot T_0^{e_0} \ldots T_n^{e_n} $$ In other words, the terms corresponding to $i_0 < \ldots < i_p$ such that $NEG(\vec{e})$ is not contained in $\{i_0 \ldots i_p\}$ are zero. The differential of the complex $\check{\mathcal{C}}^\bullet(\vec{e})$ is still given by the exact same formula as above. \medskip\noindent Suppose that $NEG(\vec{e}) = \{0, \ldots, n\}$, i.e., that all exponents $e_i$ are negative. In this case the complex $\check{\mathcal{C}}^\bullet(\vec{e})$ has only one term, namely $\check{\mathcal{C}}^n(\vec{e}) = R \cdot \frac{1}{T_0^{-e_0} \ldots T_n^{-e_n}}$. Hence in this case $$ H^q(\check{\mathcal{C}}^\bullet(\vec{e})) = \left\{ \begin{matrix} R \cdot \frac{1}{T_0^{-e_0} \ldots T_n^{-e_n}} & \text{if} & q = n \\ 0 & \text{if} & \text{else} \end{matrix} \right. $$ The direct sum of all of these terms clearly gives the value $$ \left(\frac{1}{T_0 \ldots T_n} R[\frac{1}{T_0}, \ldots, \frac{1}{T_n}]\right)_d $$ in degree $n$ as stated in the lemma. Moreover these terms do not contribute to cohomology in other degrees (also in accordance with the statement of the lemma). \medskip\noindent Assume $NEG(\vec{e}) = \emptyset$. In this case the complex $\check{\mathcal{C}}^\bullet(\vec{e})$ has a summand $R$ corresponding to all $i_0 < \ldots < i_p$. Let us compare the complex $\check{\mathcal{C}}^\bullet(\vec{e})$ to another complex. Namely, consider the affine open covering $$ \mathcal{V} : \Spec(R) = \bigcup\nolimits_{i \in \{0, \ldots, n\}} V_i $$ where $V_i = \Spec(R)$ for all $i$. Consider the alternating {\v C}ech complex $$ \check{\mathcal{C}}_{ord}^\bullet(\mathcal{V}, \mathcal{O}_{\Spec(R)}) $$ By the same reasoning as above this computes the cohomology of the structure sheaf on $\Spec(R)$. Hence we see that $H^p( \check{\mathcal{C}}_{ord}^\bullet(\mathcal{V}, \mathcal{O}_{\Spec(R)}) ) = R$ if $p = 0$ and is $0$ whenever $p > 0$. For these facts, see Lemma \ref{lemma-cech-cohomology-quasi-coherent-trivial} and its proof. Note that also $\check{\mathcal{C}}_{ord}^\bullet(\mathcal{V}, \mathcal{O}_{\Spec(R)})$ has a summand $R$ for every $i_0 < \ldots < i_p$ and has exactly the same differential as $\check{\mathcal{C}}^\bullet(\vec{e})$. In other words these complexes are isomorphic complexes and hence have the same cohomology. We conclude that $$ H^q(\check{\mathcal{C}}^\bullet(\vec{e})) = \left\{ \begin{matrix} R \cdot T_0^{e_0} \ldots T_n^{e_n} & \text{if} & q = 0 \\ 0 & \text{if} & \text{else} \end{matrix} \right. $$ in the case that $NEG(\vec{e}) = \emptyset$. The direct sum of all of these terms clearly gives the value $$ (R[T_0, \ldots, T_n])_d $$ in degree $0$ as stated in the lemma. Moreover these terms do not contribute to cohomology in other degrees (also in accordance with the statement of the lemma). \medskip\noindent To finish the proof of the lemma we have to show that the complexes $\check{\mathcal{C}}^\bullet(\vec{e})$ are acyclic when $NEG(\vec{e})$ is neither empty nor equal to $\{0, \ldots, n\}$. Pick an index $i_{\text{fix}} \not \in NEG(\vec{e})$ (such an index exists). Consider the map $$ h : \check{\mathcal{C}}^{p + 1}(\vec{e}) \to \check{\mathcal{C}}^p(\vec{e}) $$ given by the rule $$ h(s)_{i_0 \ldots i_p} = s_{i_{\text{fix}} i_0 \ldots i_p} $$ (compare with the proof of Lemma \ref{lemma-cech-cohomology-quasi-coherent-trivial}). It is clear that this is well defined since $$ NEG(\vec{e}) \subset \{i_0, \ldots, i_p\} \Leftrightarrow NEG(\vec{e}) \subset \{i_{\text{fix}}, i_0, \ldots, i_p\} $$ Also $\check{\mathcal{C}}^0(\vec{e}) = 0$ so that this formula does work for all $p$ including $p = - 1$. The exact same (combinatorial) computation as in the proof of Lemma \ref{lemma-cech-cohomology-quasi-coherent-trivial} shows that $$ (hd + dh)(s)_{i_0 \ldots i_p} = s_{i_0 \ldots i_p} $$ Hence we see that the identity map of the complex $\check{\mathcal{C}}^\bullet(\vec{e})$ is homotopic to zero which implies that it is acyclic.