Show that $d^2(w)=0$ for every $k$-form in $\mathbb{R}^n$ The question says to start showing this for 0-forms, and I did it. But how can I extend it for any k-form?
 A: A proof can for instance be found after Proposition 14.23 on page 364 of John M. Lee's Introduction to Smooth Manifolds. For completeness, let me (essentially) recall his argument.
First, let $f$ be a $0$-form on an open subset $U\subseteq \mathbb{R}^n$, i.e. a smooth function $U\rightarrow \mathbb{R}$. By definition of the exterior derivative, we then have $$d^2f=d\Big(\sum_{i=1}^n\frac{\partial f}{\partial x_i}dx_i\Big)= \sum_{i=1}^n d\Big(\frac{\partial f}{\partial x_i}dx_i\Big)= \sum_{i=1}^n\sum_{j=1}^n \frac{\partial^2 f}{\partial x_j\partial x_i}dx_j\wedge dx_i.$$
The alternating property of the wedge product then yields $$d^2f=  \sum_{1\leq i<j\leq n}\Big(\frac{\partial^2 f}{\partial x_j\partial x_i} -\frac{\partial^2 f}{\partial x_i\partial x_j}\Big)dx_j\wedge dx_i.$$
Schwarz's theorem then implies $d^2f=0.$
Next, recall the Leibniz rule for the exterior derivative.

Proposition. If $\omega$ is a smooth $k$-form and $\eta$ is a smooth $l$-form on an open subset $U\subseteq \mathbb{R}^n$, then $$d(\omega\wedge\eta)=d\omega\wedge\eta+(-1)^k\omega\wedge d\eta.$$

As Ivo Terek notes in the comments, by linearity of the exterior derivative, it suffices to prove that for any simple $k$-form we have $d^2\omega=0$. To do so, let $\omega=f dx_{i_1}\wedge\ldots \wedge dx_{i_k}$ be a simple $k$-form on an open subset $U\subseteq \mathbb{R}^n$, where $f\colon U\rightarrow \mathbb{R}$ is a smooth function. By applying the above graded Leibniz rule multiple times (formally, this amounts to an induction proof), we see
$$d^2\omega=d\Big(df\wedge dx_{i_1}\wedge\ldots \wedge dx_{i_k}\Big)=d^2f \wedge dx_{i_1}\wedge\ldots \wedge dx_{i_k}-df\wedge d\Big(dx_{i_1}\wedge\ldots \wedge dx_{i_k} \Big)=\ldots =d^2f \wedge dx_{i_1}\wedge\ldots \wedge dx_{i_k} +\sum_{j=1}^k (-1)^jdf\wedge dx_{i_1}\wedge \ldots \wedge d^2 x_{i_j}\wedge \ldots \wedge dx_{i_k}.$$
Since $d^2\eta=0$ for any $0$-form $\eta$, we in particular know that $d^2f=0$ and $d^2x_{i_j}=0$ for $1\leq j \leq k$. Thus, $d^2\omega=0.$
