How can I show that
$$ \mathrm d(a\wedge b)=(\mathrm d\,a)\wedge b + (-1)^{q}a\wedge(\mathrm d\,b) $$
for a $q$-form $a$ and an $r$-form $b$?
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2First of all, your sign is wrong on the second term! – Ted Shifrin Oct 24 '16 at 20:31
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@TedShifrin Better now? – Andy Miles Oct 25 '16 at 17:57
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1Much better. You'll find the $(-1)^q$ when you write out the proof and use the product rule. – Ted Shifrin Oct 25 '16 at 18:14
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Andy, briefly, any $q$-form can be written as $$\sum_{j_1<j_2<\dots<j_q} f_{j_1j_2\dots j_q}dx_{j_1}\wedge\dots\wedge dx_{j_q} = \sum_J f_J dx_J,$$ for short. Since $\wedge$ is bilinear and since the exterior derivative of a sum is the sum of the exterior derivatives, it suffices to take just one such term for each of $a$ and $b$ and take $$a = f_J\,dx_J \quad\text{and}\quad b = g_I\,dx_I.$$ Then $a\wedge b = f_Jg_I dx_J\wedge dx_I$ (which will be $0$, of course if there's repetition in the multiindices). Now use the product rule and rearrange terms (see note below) to get your answer. (Note that if there's repetition in these multiindices, that repetition will persist when you differentiate and so you'll get $0$ automatically on the right-hand side, too.)
(Note: $$dx_r\wedge dx_J\wedge dx_I=(-1)^q dx_J\wedge dx_r\wedge dx_I$$ for any $1\leq r\leq n$.)
Ted Shifrin
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Ah that's what you meant with "by linearity" in the comments to Danu's answer. So if I view this as a totally antisymmetric tensor (i.e. matrix), I just pick one matrix component (i.e. just one function $f$ such that $a=f \mathrm dx_I$, no indices in the function, because we fixed the component), prove it for this component, and by linearity we're done. – Andy Miles Oct 26 '16 at 07:20