Let $\Omega^k (U)$ be a set of $k-$ diffrential form on $U$, where $U \subset \mathbb R^n$ is region, and define $ Z^k (U)$ be a set of $k-$ closed form, $B^k (U)$ be a set of $k-$ exact form.
Then, de Rham cohomology $H^k (U)$ is defined as $H^k (U)=Z^k (U) / B^k (U)$.
I want to show $H^k (\mathbb R^n)= \begin{cases} \{0 \} \ &(k>0)\\ \ \mathbb R \ &(k=0)\\ \end{cases} $
Here is what I did so far.
For $k>0,$ $B^k(\mathbb R^n)=Z^k (\mathbb R^n)$ (Poincare's Lenma) thus $H^k(\mathbb R^n)=Z^k(\mathbb R^n)/ Z^k (\mathbb R^n)=\{ Z^k (\mathbb R^n) \}.$
For $k=0,$ I have $B^0 (\mathbb R^n)=\{ 0\},$ \begin{align*} H^0 (\mathbb R^n)&=\{ [\omega] \mid \omega \in Z^0 (U) \} \\ &=\{ [f] \mid f \in C^\infty (U), df=0 \}\\ &=\{ [c] \mid c\in \mathbb R \} \end{align*} where $[ \cdot ]$ is an equivalent class with $\omega \sim \omega' \iff \omega-\omega'\in B^k (U)$.
So, my question is
(i) Why $\{ Z^k (\mathbb R^n) \}=\{0 \}$ for $k>0$ ?
(ii) Why $\{ [c] \mid c\in \mathbb R \}=\mathbb R$ ?
I'd like you to give me any help for understandng this.