Part of proof that $d^2\omega=0$ The following comes from the proof in differentiable manifolds that $d^2\omega=0$.
Let $f$ belong to the set of $0$-forms. From definition I have that 
$\displaystyle df = \frac{\partial f}{\partial x^j}dx^j$
Then from the definition for $d$ of a one-form, apparently we have that
$\displaystyle d(df)=d\Big(\frac{\partial f}{\partial x^j}\Big) \wedge dx^j$.
I cannot see how this can been derived. 
I tried $\displaystyle d(df)=d\Big(\frac{\partial f}{\partial x^j} dx^j \Big)$. You then use some kind of product rule?
The definition I have been given is if 
$\displaystyle \omega = \frac{1}{k!} \omega_{i_1 ... i_k} dx^{i_1}\wedge...\wedge dx^{i_k}$ then its derivative is
$\displaystyle d\omega = \frac{1}{k!} d\omega_{i_1\cdots i_k} \wedge dx^{i_1}\wedge\cdots\wedge dx^{i_k}$

I'll write out an answer for a form of three variables.  This might yield some intuition.  Thing to notice is that $dx^i\wedge dx^i=0$, and that $dx^i\wedge dx^j=-dx^j\wedge dx^i$, and $f_{xy}=f_{yx}$.
Thus, because $dx^i\wedge dx^i=0$
$$d^2f=f_{xy}dy\wedge dx+f_{xz}dz\wedge dx+f_{yx}dx\wedge dy+f_{yz}dz\wedge dy+f_{zx}dx\wedge dz+f_{zy}dy\wedge dz   $$
Do you see how these terms cancel out to give $0$?
