Exterior power of cotangent bundle of a cartesian product of manifolds

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Let $M \times F$ be a product manifold. The identity $T^*(M \times F) = T^*(M) \times T^*(F)$ holds generally (I hope). What can be said about the exterior powers of it?

$$\bigwedge^r T^*(M \times F)\cong \space?$$

Motivation:

Let $P \to M$ be a fiber bundle with fiber $F$. let $\omega \in \bigwedge^r T^*P$. Is it possible in general to factor $\omega$ locally as a product of basic forms? Explicitly: given a $p \in M$, does there always exist a neighbourhood $p\in U$ s.t. $\omega|_U$ is a sum of simple forms of the type $\nu \wedge \alpha$ for some basic forms $\nu \in \bigwedge^{r-q} M$ and $\alpha \in\bigwedge^{q} F$?

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The cotangent bundle of $M \times F$, as a vector bundle, is the direct sum of the cotangent bundles of $M$ and $F$ (more precisely, of their pullbacks along the natural projections), so you're asking how to describe the exterior powers of a direct sum $V \oplus W$ of two vector bundles. The answer is that the exterior algebra is a graded tensor product

$$\bigwedge^{\bullet}(V \oplus W) \cong \bigwedge^{\bullet}(V) \otimes \bigwedge^{\bullet}(W)$$

and this gives a "Kunneth formula" for the individual terms, namely

$$\bigwedge^k(V \oplus W) \cong \bigoplus_{i+j = k} \bigwedge^k V \otimes \bigwedge^k W.$$

Abstractly, the exterior algebra functor is left adjoint to the forgetful functor from graded commutative algebras to vector spaces given by taking the degree $1$ part, so it sends coproducts to coproducts.

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$\bigwedge^r T^*(M\times F)=\sum_{s+t=r}\bigwedge^s T^*M\otimes\bigwedge^t T^*F$