Prove $d(f\alpha)=d(f \wedge \alpha)$ I am reading the article http://en.wikipedia.org/wiki/Exterior_derivative and a definition of an exterior derivative from Axioms for the exterior derivative. How could I show that if $f$ is a function ($0$-form) and $\alpha$ a $k$-form, then $d(f\alpha)=d(f \wedge \alpha)$?
 A: Functions are 0-forms, I.e., scalars. Scalar multiplication and the exterior product are equivalent when one of the arguments is a scalar, which is why $ d (f\alpha)=d (f\wedge\alpha) $.
A: The multiplication $f\alpha$ is just scalar multiplication.  Explicitly, let $e^{p,1}, e^{p,2}, \ldots$ be a basis of $p$-forms.  Then the $p$-form $\alpha$ can be written in terms of $\binom{n}{p}$ scalar fields $\alpha_1, \alpha_2, \ldots$ like so:
$$\alpha = \sum_{i=0}^{\binom{n}{p}} \alpha_i e^{p,i}$$
The functions $\alpha_i$ are functions of coordinates. That is, $\alpha_i = \alpha_i(x^1, x^2, \ldots, x^n)$.  The basis $p$-forms $e^{p,i}$ are functions of both coordinates and $p$ input vectors, but the functional dependence on the input vectors isn't really relevant or necessary to consider when just taking an exterior derivative.  That's all linear, and not very interesting.  I'll just abstract that out as some $p$-vector $A$, so we can write $f\alpha$ like so:
$$f(x)\alpha(x, A) = \sum_{i=0}^{\binom{n}{p}} f(x) \alpha_i(x) e^{p,i}(x,A)$$
You can just define a new $p$-form $\beta$ such that $\beta_i = f \alpha_i$.  The wedge product of a scalar and a $p$-form is then identified with scalar multiplication in this manner. 
A: Here is how it's done in Madsen and Tornehave's book "From calculus to cohomology":
For an alternating $p$-form $\omega$ and an alternating $q$-form $\tau$, the alternating  $p+q$-form $\omega \land \tau$ is defined by
$$(\omega \land \tau)(v_1, \ldots, v_{p+q}) = \sum_{\sigma \in S(p,q)} \text{sgn} (\sigma)\omega(v_{\sigma(1)}, \ldots, v_{\sigma(p)})\tau(v_{\sigma(p+1)}, \ldots, v_{\sigma(p+q)})$$
where $S(p,q) \subseteq S_{p+q}$ is the set of $(p,q)$-shuffles.
If $\omega$ is an alternating 0-form, we get
$$(\omega \land \tau)(v_1, \ldots, v_{p+q}) = \sum_{\sigma \in S(0,q)} \text{sgn}(\sigma)\omega(\cdot)\tau(v_{\sigma(1)}, \ldots, v_{\sigma(q)}) = \omega \tau(v_1, \ldots, v_p)$$
So $\omega \land \tau = \omega \tau$.
When we move to differential forms, with $f$ a differential 0-form and $\alpha$ a differential $q$-form, we get
$$(f\land \alpha)(x) = f(x) \land \alpha(x) = f(x) \alpha(x)$$
since $f(x)$ is an alternating 0-form and $\alpha(x)$ is an alternating $q$-form. Thus $f \land \alpha = f \alpha$.
