Positive-degree cohomology of a compact manifold is nilpotent

For $F\mathrm{\ne }\mathrm{\varnothing }$, choose the finite cover $U_1,\dots ,U_N$ together with its nerve homotopy equivalence supplied by [F4], with $N\ge 1$. Every nonempty finite intersection $U_{i_1}\cap \cdots \cap U_{i_s}$ is strongly geodesically convex. Choosing a point in that intersection, the unique minimizing geodesics to it give a continuous contraction to the point; by [F2] and the singular cochain definition in [F3], its cohomology is $k$ in degree zero and zero in positive degrees. For a point, this follows from the unique singular $n$-simplex: its boundary coefficient is $1$ for even $n$ and $0$ for odd $n$, so the positive-degree cochain complex is alternately identity and zero and has zero cohomology. The nerve of this same cover has at most $N$ vertices; its usual realization, with one open cell for each nonempty simplex, is a finite CW complex homotopy equivalent to $F$ by the chosen [F4] data.

We prove by induction on the number $m\ge 1$ of members of a finite good cover that its union has cohomology zero in every degree $q\ge m$. For $m=1$, step 1.1 makes the member contractible, giving the claim. Empty cover members have zero cohomology by [F3] and may be discarded; if none remain the union is empty and the assertion holds. For $m>1$, write $A=U_1\cup \cdots \cup U_{m-1}$ and $B=U_m$. The intersection $A\cap B$ is covered by the $m-1$ sets $U_i\cap U_m$, whose nonempty finite intersections are contractible by step 1.1. The induction hypothesis gives $H^q(A;k)=0$ and $H^{q-1}(A\cap B;k)=0$ for $q\ge m$; also $H^q(B;k)=0$ since $q>0$. The exact segment $H^{q-1}(A\cap B;k)\to H^q(A\cup B;k)\to H^q(A;k)\oplus H^q(B;k)$ from [F3] therefore forces $H^q(A\cup B;k)=0$.

Apply step 2.1 to the selected finite cover of $F$ to get $H^q(F;k)=0$ for $q\ge N$. Every homogeneous factor of a product of $N$ elements of $J$ has positive degree, so the product has degree at least $N$ and vanishes; expanding finite sums proves $J^N=0$. For empty $F$, every cohomology group is zero by the empty-space cochain complex, so $N=1$ works.