Proving the de Rham cohomology of $M$ and $M \times \mathbb R$ are isomorphic without homotopy Given a smooth manifold $M$, is there an elementary way of showing $H^k_{dR} (M) \cong H^k_{dR} (M \times \mathbb R)$, i.e. straight from the definitions of cohomology being the quotient of closed forms by exact forms, without homotopy or integration on $M$? 

The idea I had was to write forms on $M \times \mathbb R$ in terms of those on $M$, however it is not entirely obvious to me how to go about this and convert to a statement about cohomology.
The existing answer is worthless! Forms on $\Bbb R \times M$ do not take the shape "$dt \wedge \alpha$", they are much more complicated. So nothing has been proved.

A $k$-form on $N = \Bbb R \times M$ takes the shape $dt \wedge \alpha_t + \beta_t$, where $\beta_t$ is a time-varying $k$-form on $M$ and $\alpha_t$ is a time-varying $(k-1)$-form. 

Then $$d_N(dt \wedge \alpha_t + \beta_t) = dt \wedge (\beta'_t - d_M \alpha_t) + d_M \beta_t.$$

For our form to be closed, then, we would have $\beta'_t = d_M \alpha_t$ and $d_M \beta_t = 0$. Thus we may write $\beta_t = \beta_0 + \int_0^t d_M \alpha_s ds$, where $d_M \beta_0 = 0$. 

Claim: suppose $\beta_0 = d_M \eta_0$. Then what we seek are time-varying forms $\gamma_t$ and $\eta_t$ with $\eta'_t - d_M \gamma_t = \alpha_t$ and $\beta_t = d_M \eta_t.$ This is simple enough: take $\eta_t = \eta_0 + \int_0^t \alpha_s ds.$ Now we have the equation $d_M \gamma_t = \eta'_t - \alpha_t$; now one needs to know that there is a left-inverse $A$ to $d_M$ on the space of exact forms. This would give $\gamma_t = A(\eta'_t - \alpha_t)$. 

Usually this operator $A$ comes from Hodge theory, which says that $\text{ker}(d_M^*) \xrightarrow{d_M} \text{im}(d_M)$ is surjective with finite-dimensional kernel; here $d_M^*$ is the adjoint operator. Take the existence of this operator $A$ for granted.

What we have shown is that every closed form with $\beta_0$ exact is in fact an exact form. This means that the map $H^k(M \times \Bbb R) \to H^k(M)$ given by restricting to the time-0 slice is injective. But surjectivity is clear because if $\alpha$ is a closed form on $M$ then $dt \wedge \alpha$ is a closed form on $M \times \Bbb R$.
Hint: Let $\alpha$ be a form of $M$, suppose that $dt\wedge \alpha$ is closed, it implies that $\alpha$ is closed, $d(t.\alpha)=dt\wedge\alpha$