I apologise that this is similar to older questions, but I’ve seen no definitive answer yet. If $a,b,c$ are $1$-forms and $b\otimes c$ is the tensor product, is the wedge product $a\wedge ( b\otimes c)$ defined or is it just meaningless?
I ask because I was trying to derive the associative law for the wedge product, i.e., $a∧(b∧c)=(a\wedge b)\wedge c=a\wedge b\wedge c$ starting with the definition $b\wedge c=b\otimes c-c\otimes b$. One then has to make sense of $a\wedge (b\otimes c-c\otimes b)$.
I made the assumption that this must be,
$a\wedge(b\otimes c-c\otimes b)≡a\otimes(b\otimes c-c\otimes b)-(b\otimes c-c\otimes b)\otimes a$
But this is just wrong because it does not produce the totally antisymmetric tensor that a∧b∧c actually is. So that leaves me with the question, how to make sense of $a\wedge (b\otimes c-c\otimes b)$, so it gives the right answer – or am I on a fool’s errand because wedge products like $a\wedge( b\otimes c)& are just not defined? That’s the only solution I can think of.
You are using the differential geometer's definition of wedge products in terms of antisymmetric tensors, which requires many factorials in coefficients to make things associative. A proof using your definition is Theorem 4-4(3) of Spivak's Calculus on Manifolds and Prop. 12.8 of Lee's Introduction to Smooth Manifolds. Without those factorials, associativity breaks down: see the paragraph after Lemma 2.2 and the last page here.
If you define exterior powers of a vector space as a quotient space of a tensor power rather than as a subspace, then associativity of wedge products is less messy to show: see Definition 3.1 and Theorem 8.3 here, where you can replace the generality of modules over a commutative ring with a special case for your purposes: finite-dimensional real vector spaces.