= Solution
The <de Rham cohomology> is
$$
H^p_{\mathrm{dR}}(M)=\ker(d:\Omega^p\to\Omega^{p+1})/operatorname{im}(d:\Omega^{p-1}\to\Omega^p).
$$
The <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 $B'$ around $B$. The Poincare lemma gives $\alpha=dh$ on $B'$. The annulus $B'\setminus B$ 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 put
$$
A(x,t)=\int_{t_0}^t a(x,s)\,ds.
$$
Since $d_{S^n}A\wedge\omega=0$ for dimensional reasons,
$$
d((-1)^nA\omega)=a\omega\wedge dt.
$$
Every top-degree form is exact, so $H^{n+1}(S^n\times I)=0$ without using the de Rham theorem.
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