Let $f$ be a differentiable function on an open space $U \subset \mathbb{R}^n$. Proove that $d(df) = 0$.
So my proof is:
Let
$$f = \sum c_{i_1, \cdots i_k}(x_*)dx_{i_1} \wedge \cdots dx_{i_k}$$
where $c_{i_1, \cdots i_k}$ are constants and $x_* = (x_1, \cdots x_n)$. We then get
$$df = \sum d[c_{i_1, \cdots i_k}(x_*)] \wedge dx_{i_1} \wedge \cdots dx_{i_k}$$ $$\implies d(df) = \sum d\{ d[c_{i_1, \cdots i_k}(x_*)] \wedge dx_{i_1} \wedge \cdots dx_{i_k}\} = 0.$$
How does that last line $=0$ though?
You seem to be trying to prove it for $k$-forms. It also works and the proof is basically the same. What you say in the comment is correct. If $f$ is a smooth function then $df=\sum\limits_{i=1}^n\frac{\partial f}{\partial x_i}(x)dx^i$. So when you take $d$ again you get $d(df)=\sum\limits_{i,j=1}^n\frac{\partial^2 f}{\partial x_j \partial x_i}(x)dx^j\wedge dx^i.$ Now you only need to write this in terms of $\{dx^k\wedge dx^l:k<l\}$ which is the standard basis for $2$-forms.
Hint: Recall that $dx^i\wedge dx^i=0$ and $dx^i\wedge dx^j=-dx^j\wedge dx^i$.