Prove the exterior derivative of the following (n-1) form is zero Let $\omega(x)=\frac{1}{{\parallel x \parallel}^n}\displaystyle\sum_{i=1}^{n}(-1)^{i-1}x_{i} dx_{1} \wedge \dots \wedge \widehat{dx_{i}} \wedge \dots \wedge dx_{n}$ be a differential $(n-1)$ form on $\mathbb{R}^n \setminus\{0\}$. Where $\widehat{dx_{i}}$ means that the $i$th term is absent from the product. Prove that the exterior derivative $d\omega=0$.
What I have so far is $d\omega(x)=\displaystyle\sum_{i=1}^{n}d(\frac{1}{{\parallel x \parallel}^n})(-1)^{i-1}dx_{i}\wedge dx_{1} \wedge \dots \wedge \widehat{dx_{i}} \wedge \dots \wedge dx_{n}$
$=\displaystyle\sum_{i=1}^{n}d(\frac{1}{{\parallel x \parallel}^n})(-1)^{i-1} (-1)^{i}dx_{1} \wedge \dots \wedge dx_{n}$
$=\displaystyle\sum_{i=1}^{n}d(\frac{1}{{\parallel x \parallel}^n})dx_{1} \wedge \dots \wedge dx_{n}$
However I am unsure on how to proceed from here. I'm also not sure if what I have done so far is even correct. Any help would be greatly appreciated. 

If $$\omega = \sum_{i=1}^n\frac{(-1)^{i-1}x_i}{\|x\|^n}\,dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n,$$then: $$d\omega = \sum_{i=1}^n\sum_{j=1}^n\frac{\partial}{\partial x_j}\left(\frac{(-1)^{i-1}x_i}{\|x\|^n}\right) dx_j \wedge dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n.$$Now, the only surviving term is when $j = i$, and we'll need $i-1$ inversions to write $dx_i\wedge dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n$ as $dx_1 \wedge \cdots \wedge dx_n$. Meaning: $$d\omega = \left(\sum_{i=1}^n \frac{\partial}{\partial x_i}\left(\frac{x_i}{\|x\|^n}\right)\right) dx_1 \wedge \cdots \wedge dx_n.$$
Now we compute: \begin{align}\frac{\partial}{\partial x_i}\left(x_i\left(\sum_{j=1}^n x_j^2\right)^{-n/2}\right) &= \left(\sum_{j=1}^n x_j^2\right)^{-n/2} - \frac{n}{2}x_i \left(\sum_{j=1}^n x_j^2\right)^{-n/2-1} (2x_i) \\ &= \frac{1-n\frac{x_i^2}{\|x\|^2}}{\|x\|^n} \end{align}
Attacking the above expression with $\sum_{i=1}^n$ gives $$\sum_{i=1}^n \frac{\partial}{\partial x_i}\left(\frac{x_i}{\|x\|^n}\right) = \frac{n - n\frac{\|x\|^2}{\|x\|^2}}{\|x\|^n} = \frac{n-n}{\|x\|^n}=0.$$Hence $d\omega = 0$.

Your computation has a few mistakes. For example, you shouldn't write $$\cdots\sum_{i=1}^{n}d(\frac{1}{{\parallel x \parallel}^n})(-1)^{i-1}dx_{i} \cdots$$because you're taking the differential of $(-1)^{i-1}x_i/\|x\|^n$, which is different from the differential of $(-1)^{i-1}/\|x\|^n$ times $dx_i$.
I'll leave you to mimic what I've done above to prove that if $F = (F_1,\cdots,F_n)$ is a vector field and $$\omega = \sum_{i=1}^n (-1)^{i-1}F_i\,dx_1\wedge \cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n,$$then $d\omega = ({\rm div}\,F)\,dx_1\wedge\cdots\wedge dx_n$.
