Let $\phi $ be an exterior $k$-form, where $k$ is an odd integer. Show that $\phi \wedge \phi =0 $

Question

Let $\phi $ be an exterior $k$-form, where $k$ is an odd integer. Show that $\phi \wedge \phi =0 $

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We know that If $\phi$ is a $k$ form and $\pi $ is a $l$ form then $\phi \wedge \pi = (-1)^{kl} \pi \wedge \phi$.
By the given condition $k$ is odd . So $k^2$ is also odd.
Now $\phi \wedge \phi =(-1)^{k^2}\phi \wedge \phi $.
Since $k^2$ is odd we have $\phi \wedge \phi =-\phi \wedge \phi $.
therefore $\phi \wedge \phi =0 $.

Am I right?

Answer 1

We know that If $\phi$ is a $k$ form and $\pi $ is a $l$ form then $\phi \wedge \pi = (-1)^{kl} \pi \wedge \phi$.
By the given condition $k$ is odd . So $k^2$ is also odd.
Now $\phi \wedge \phi =(-1)^{k^2}\phi \wedge \phi $.
Since $k^2$ is odd we have $\phi \wedge \phi =-\phi \wedge \phi $.
therefore $\phi \wedge \phi =0 $.