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Let $\mathcal{F}$ be any quasi-coherent sheaf on $X$, and choose $m$ from (i). Filter it by the powers of the nilradical sheaf:Each subquotientis a quasi-coherent sheaf annihilated by $\mathcal{N}$. Apply (ii) to the quasi-coherent sheaf ${\mathcal{N}}^j\mathcal{F}$. Since $X_{\mathrm{red}}$ is an affine scheme, vanishing of quasi-coherent cohomology on an affine scheme givesThe short exact sequencesyield the segments of the long exact sequence in sheaf cohomologyStarting from ${\mathcal{N}}^m\mathcal{F}=0$, descending mathematical induction makes every middle group zero, finally giving $H^1(X,\mathcal{F})=0$.

This holds for every quasi-coherent sheaf $\mathcal{F}$. Since a Noetherian scheme is a quasi-compact scheme, the Serre criterion for affineness in (a) applies and provesThis is an instance of the general fact that affineness is invariant under nilpotent thickening.

A Noetherian scheme is a quasi-compact scheme, so choose a finite affine open cover $X={\bigcup }_{i=1}^s\mathrm{Spec}{A}_i$ with all $A_i$ Noetherian rings. The nilradical sheaf restricts on this chart to the quasi-coherent sheaf associated to $\mathrm{Nil}(A_i)$: the nilradical commutes with localization.

By the allowed fact, choose $d_i\ge 1$ such that $\mathrm{Nil}(A_i{)}^{d_i}=0$ and put $m={\max }_i{d}_i$. On every chart, the ideal sheaf ${\mathcal{N}}^m$ is zero, and hence it is zero globally. Higher powers are also zero. ThusIf $X$ is empty, take $m=1$. The finite affine open cover is essential for choosing one bound valid everywhere.