Exterior derivative of $(dx^{i_1} \land ... \land dx^{i_p})$

Question

I am currently working on differentialforms. We introduced the exterior derivative, as follows:

The exterior derivative is the operator $d: \Lambda^p \rightarrow \Lambda^{p+1}$ with the properties:

$$1. \quad d( \alpha + \beta ) = d\alpha + d\beta$$ $$2. \quad d^2 = 0$$ $$3. \quad \text{on 0-forms}: df = \tfrac{\partial f}{\partial x^i} dx^i$$ $$4. \quad \text{for a function $f$ and a form $\omega$}: d(f\omega) = (df) \land \omega + f d\omega$$

Now, one of the problems I have to work on is the following:

Show that $d( dx^{i_1} \land ... \land dx^{i_p} ) = 0$.

I am afraid, that I have no clue on how to show this. Can anyone provide a hint or a solution?

Answer 1

Hint: Try to proceed by induction on $p$. Start with $d(dx^i)=0$ (property 2), use this to show that for $i<j$ you get $d(x^idx^j)=dx^i\wedge dx^j$, which implies the statement for $p=2$ and so on.