$H^{n}(M)$ where $M$ is compact, orientable and connected manifold I need to show that if $M$ is compact, orientable and connected manifold of dimension $n$, then $H^{n}(M) = \mathbb{R}$. 
I saw that a possible proof is to take an atlas for $M$, with $U_{\alpha}$, $\alpha = 1,..N$ and $\rho_{\alpha}$ be a smooth partition of unity. Then define a smooth map 
\begin{equation}
\xi: \Omega^{n}(M) \rightarrow \mathbb{R}^{N}
\end{equation}
by 
\begin{equation}
\xi(\omega) = (\int_{M}\rho_1\omega,..,\int_{M}\rho_N\omega)
\end{equation}
and consider the subspace $X = \{\xi(dv) | v \in \Omega^{n-1}(M)\}$.
The proof than shows that (proof is omitted) 
(A) $\omega$ is exact if and only if $\xi(\omega) \in X$
(B) $w \in \Omega^{n}(M)$ is exact if and only if $\int_M\omega=0$.
I have two questions: 
1) What does it means $\omega$ is exact? It's not a sequence, just an element of $\Omega^{n}(M)$.
2) Why (B) actually implies that $H^{n}(M) = \mathbb{R}$ ?
Thanks

*

*A differential $n$-form $\alpha$ is exact if there is some differential $(n-1)$-form $\beta$ such that $d\beta = \alpha$. 

*Recall that $H^n(M)$ is defined by taking the closed $n$-forms and modding out by the exact $n$-forms. Fact (B) implies that integration gives you a well-defined map $H^n(M)\rightarrow \mathbb R$, since any exact form integrates to zero. In fact, you can show that this is a vector space isomorphism. (Hint: Take a non-zero closed form $\alpha$ and show that any other closed form is equivalent, modulo some exact form, to a multiple of $\alpha$. Use the integration map.)
(Note that part $B$ follows from Stokes's theorem.)
