If $f$ is a $0$-form and $dx$ a $1$-form, then does $f\ dx=f\wedge dx$?
Background: I am trying to prove that $d(f\ dx)=df\wedge dx$ using only the four properties of $d$:
(1) $d(w+v)=dw+dv$
(2) $d(w\wedge v)=dw\wedge v+(-1)^{\deg w}w\wedge dv$
(3) $f$ is a $0$-form $\implies df$ is the differential of $f$
(4) $f$ is a $0$-form $\implies d^2f=0$
My proof: $d(f\ dx)=d(f\wedge dx)=df\wedge dx+f\wedge d^2x=df\wedge dx$. The second and third equal signs come from property 2 and 4 respectively. But the first one must come from $f\ dx=f\wedge dx$. My question is why this is true.
Although it seems very trivial, I believe a rigorous explanation is as follows: Let $A$ be the alternating map from the tensor product of $0$-forms and $1$-forms to its subspace of alternating tensors. Then by definition $f\wedge dx=A(f\otimes dx)$. But there is a multiplication defined between $0$-forms and $1$-forms, so that tensoring means multiplying, and $f\otimes dx= f\ dx$. And $A$, as the alternating map over $1$-forms, is actually the identity. Hence $f\wedge dx=A(f\otimes dx)=f\ dx$
Is this a correct explanation?