property of the exterior derivative $d \circ d=$ for a $\mathcal C^\infty$ function One of the properties of the exterior derivative is that $d\circ d=0$. We're trying to prove this for the case $f\in\mathcal C^\infty (U)$ on an open set $U\subset \mathbb R^n $.
The prove starts with the uniqueness of the exterior derivative using the properties:


*

*$d$ is $\mathbb R$-linear

*$d\circ d=0$

*If $f$ is in $\mathcal C^\infty(U)$, $df$ is the differential of $f$.

*$d(\omega\wedge\eta)=d\omega\wedge \eta+(-1^k) \omega \wedge d\eta$


Now to the existence of $d$: 
What I'm interested in is the proof of $d\circ d=0$. Let $f\in\mathcal C^\infty (U)$.
$$
\begin{eqnarray}
ddf &=& d(\sum_{i=0}^n \frac{\partial f}{\partial x_i} dx_i) \\ &=&
\sum_{i=0}^n d(\frac{\partial f}{\partial x_i}) \cdot  dx_i+ \sum_{i=0}^n (\frac{\partial f}{\partial x_i}) d(dx_i)
\end{eqnarray}
$$
The argument for the left part of the sum is the lemma of schwartz and the alternating property and i got the soultion.The other part is the problem.
Now $d(dx_i)$ is the exterior derivative of a 1-form, as were trying to prove $d\circ d=0$ it is not clear to me on how to deduce that this part of the sums is $0$.
 A: In your list of properties you should add locality, i.e. if 2 forms $\alpha$ and $\beta$ coincide on an open set $U$, then $d\alpha=d\beta$ on $U$.
More generally, denoting by $M$ a given manifold, if $\omega\in\Omega^{n}(M)$ is locally represented by $w=w_{i_1\dots i_n}dx_{i_1}\wedge\dots\wedge dx_{i_n}$, then, denoting by $d'$ an operator that satisfies all properties 1-4 above, by using 4)
$$d'w=(d'w_{i_1\dots i_n})dx_{i_1}\wedge\dots\wedge dx_{i_n}+
w_{i_1\dots i_n}d'(dx_{i_1}\wedge\dots\wedge dx_{i_n});
 $$
using 3) we arrive at $ d'w_{i_1\dots i_n}=dw_{i_1\dots i_n}$ and $d'x_{i_l}=dx_{i_l}$.
Using 3) and 4) we have then $d'(dx_{i_1}\wedge\dots\wedge dx_{i_n})=0$, as $d'x_{i_l}=dx_{i_l}$ for all $l=1,\dots, n$.
So $d'$ is the uniquely defined operator $d$, where $d$ is the operator 
$$dw=\frac{\partial w_{i_1\dots i_n}}{\partial x_k}dx_k\wedge dx_{i_1}\wedge\dots\wedge dx_{i_n}.$$
Property 2) is satisfied as 
$$ d(dw)=\frac{\partial^2 w_{i_1\dots i_n}}{\partial x_k\partial x_r}dx_r\wedge dx_k\wedge dx_{i_1}\wedge\dots\wedge dx_{i_n}=0,$$
as we sum over all $r,k=1,\dots, n$, with $dx_r\wedge dx_k=-dx_k\wedge dx_r$.
