[proofplan] The theorem follows from the Bochner-Kodaira-Nakano identity. Dolbeault's theorem identifies the sheaf cohomology group with harmonic $(n,q)$-forms valued in $L$. Positivity of the curvature term forces every such harmonic form with $q\ge1$ to have zero norm, so the cohomology group vanishes. [/proofplan] [step:Translate sheaf cohomology to harmonic forms] By the Dolbeault isomorphism, \begin{align*} H^q(X,\Omega_X^n\otimes L)\cong H^{n,q}_{\bar\partial}(X,L). \end{align*} Since $X$ is compact Kähler, Hodge theory identifies each Dolbeault cohomology class with a unique harmonic $L$-valued $(n,q)$-form. It is therefore enough to prove that every harmonic $L$-valued $(n,q)$-form is zero for $q\ge1$. [/step] [step:Apply the Bochner-Kodaira-Nakano identity] Let $\alpha$ be a harmonic $L$-valued $(n,q)$-form. Harmonicity means \begin{align*} \bar\partial\alpha=0,\qquad \bar\partial^*\alpha=0. \end{align*} The Bochner-Kodaira-Nakano identity gives \begin{align*} 0=\|\bar\partial\alpha\|^2+\|\bar\partial^*\alpha\|^2 =\|\nabla''\alpha\|^2+\langle [i\Theta(L),\Lambda]\alpha,\alpha\rangle. \end{align*} Here $\Theta(L)$ is the Chern curvature of $L$, and $\Lambda$ is contraction with the Kähler form. [/step] [step:Use positivity of $L$] The line bundle $L$ has positive curvature, so the curvature operator $[i\Theta(L),\Lambda]$ is positive definite on $(n,q)$-forms for every $q\ge1$. Hence there is a constant $c>0$ such that \begin{align*} \langle [i\Theta(L),\Lambda]\alpha,\alpha\rangle\ge c\|\alpha\|^2. \end{align*} Substituting into the Bochner identity gives \begin{align*} 0\ge c\|\alpha\|^2. \end{align*} Thus $\alpha=0$. [/step] [step:Conclude vanishing] Every cohomology class in $H^q(X,\Omega_X^n\otimes L)$ has a harmonic representative, and the preceding step shows that this representative is zero when $q\ge1$. Therefore the cohomology class itself is zero. Hence \begin{align*} H^q(X,\Omega_X^n\otimes L)=0\qquad(q\ge1), \end{align*} as claimed. [/step]