show that the compactly supported De Rham cohomology groups $H^{p}_{DR}(\mathbb{R^n})$ are all zero for $0\leq pI am reading John Lee's book and there is a problem on De Rham cohomology: show that the compactly supported De Rham cohomology groups $H^{p}_{DR}(\mathbb{R^n})$ are all zero for $0\leq p<n$.
My confusion here lies what is the relationship between compactly supported cohomology and "normal" De Rham cohomology. 
Since $H^{n}(M)\neq0$, so I can't just use the fact that $\mathbb{R^n}$ is contractible?
 A: Compactly supported cohomology is the cohomology of the complex of compactly supported differential forms (i.e., smooth forms that are zero outside of some compact set). Let's denote compactly supported cohomology and the usual de Rham cohomology by $H^k_c(M)$ and $H^k_{DR}(M)$, respectively.
Note that on compact manifolds, every differential form is compactly supported, so the two notions of cohomology are the same. But on non-compact manifolds, they can be very different. For example, $H^0_{DR}(R^n)$ is generated by the (cohomology class) of the constant function 1. But 1 is not compactly supported, so it isn't allowed in compactly supported cohomology. In fact, $H^0_c(R^n) = 0$. (As a commenter above says, another difference between the two is that compactly supported cohomology is not a homotopy invariant in general.)
An important relationship between the two on orientable manifolds $M$ is given by Poincare duality, which says that
$$H^k_{DR}(M) \cong (H^{n-k}_c (M))^\ast$$
where $n$ is the dimension of $M$ and $\ast$ denotes the vector space dual. In particular this means that (at least if everything is finite-dimensional) $H^k_{DR}(M) \cong H^{n-k}_c (M)$. So compactly supported cohomology is the same as de Rham cohomology, but "flipped."
For example, $H^n_c(R^n) \cong H^0_{DR}(R^n) \cong R$, and $H^k_c(R^n) = 0$ for $k < n$.
One could probably also compute this directly from the definition of compactly supported cohomology without using Poincare duality (which may be more in the spirit of your problem).
