# Exterior derivative of a form and $d(d\omega)=0$?

We know that in differential geometry, $d^2\omega=0$, where $\omega$ is a form and $d$ is the exterior derivative.

However if this form happens to be the exterior derivative of another form $\theta$ such that $\omega =d\theta$ then won't $d\omega$ always be zero since we could also write $d\omega =d(d\theta)$?

Forgive me if I'm missing something elementary!

That's correct! We say a form $\omega$ is closed if $d\omega = 0$, and we say that $\omega$ is exact if $\omega = d\eta$ for some form $\eta$. Your remark says, in this terminology, that every exact form is closed.