 Solution

The de Rham cohomology isThe Poincare lemma says that every closed positive-degree differential form is locally exact, and is exact on every star-shaped open subset of Euclidean space.

Let $n>1$ and let $\alpha$ be a closed one-form on $M$. The hypothesis gives $\alpha =dg$ on $M\setminus B$. Choose a slightly larger coordinate ball $B^{\prime }$ around $B$. The Poincare lemma gives $\alpha =dh$ on $B^{\prime }$. The annulus $B^{\prime }\setminus B$ is connected when $n>1$, so $d(g-h)=0$ there and $g-h$ is constant. Adjusting $h$ by this constant makes $g$ and $h$ agree on the overlap, and they glue to a global primitive of $\alpha$. Hence $H^1(M)=0$. For $n=1$ the claim fails: remove a closed proper interval from $S^1$. Its complement is an interval and has vanishing first de Rham cohomology, whereas $H^1(S^1)\cong \mathbb{R}$.

Finally choose a nowhere-vanishing $n$-form $\omega$ on $S^n$ and a coordinate $t$ on the finite interval $I$. Every $(n+1)$-form on $S^n\times I$ is $a(x,t)\omega \wedge dt$. Fix $t_0\in I$ and putSince $d_{S^n}A\wedge \omega =0$ for dimensional reasons,Every top-degree form is exact, so $H^{n+1}(S^n\times I)=0$ without using the de Rham theorem.

Solved by gpt-5.6-sol high.