deRham cohomology of a manifold with covering space $S^{n}$

(A qual problem) Let $\pi:S^{n}\rightarrow M$ be a covering map, $M$ being an orientable manifold. Show that $H_{deR}^{k}(M)=0$ for $1\leq k < n$.

We can show $H_{deR}^{1}(M)=0$ by the following argument. For a closed 1-form $\omega$ on $M$, $\pi^{*}\omega$ is closed on $S^{n}$ thus exact, so $\pi^{*}\omega$ is equal to some $df$, $f$ being a smooth function on $S^{n}$. For any deck transformation $g$ on $S^{n}$, $g^{*}\pi^{*}\omega=\pi^{*}\omega$, thus we deduce that $d(g^{*}f-f)=0$, so $g^{*}f-f=C$ for some constant $C$. Then we apply $g^{*}$ on $g^{*}f-f=C$, we get` $(g^{2})^{*}f-g^{*}f=C$. Continue applying $g^{*}$, and noting that $g$ is of finite order, say $m$, we finally get $f-(g^{m-1})^{*}f=C$. Adding these $m$ equalities together, we get $C=0$. Thus $f$ is deck transformation invariant, then $h=f\circ \pi^{-1}$ is well defined on $M$ and $\omega=dh$. Thus $\omega$ is exact.

This argument only uses the fact that $S^{n}$ is of $H_{deR}^{1}$ zero, and compact(to guarantee that $g$ is of finite order). I try to use induction for higher order deRham cohomology, but I seems to meet an essential obstacle. What special structures of $S^{n}$ should we use to carry on the argument, or is $M$ being orientable playing a role here?

I am also interested in examples(other than $\mathbb{R}\mathbb{P}^{n}$) or even a classification of manifolds with covering space $S^{n}$.

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A covering map $f\colon X \to Y$ is a local diffeomorphism; a form on $X$ around a point $p$ gives a form on $Y$ around $f(p)$. If you have a form on $X$, for each point $q$ of $Y$ sum over the form you get from each point of $f^{-1}(q)$; the forms you get glue together to a global form on $Y$. – Angelo Mar 22 '13 at 8:18
Ah, I am so stupid! So the result has nothing to do with orientability of $M$, only the finiteness of the covering. I guess there are similar results for the case when the fibre is some compact Lie group with Haar measure for which the summing(integration) also works. I will look it up. Thank you for your patience! – Yunfeng Mar 22 '13 at 8:50