Taking the exterior derivative of a 0-form I'm attempting to show that $dg(\vec{x})=\alpha$, where $\alpha=\Sigma^n_{i=1}f_idx_i$ and $g(\vec x)={1\over{p+1}}\Sigma_{i=1}^nx_if_i(\vec x)$...and $d$ is the exterior derivative.  The $f_i$ are all smooth and homogeneous of the same ($p\neq-1)$) degree, i.e. $f_i(t\vec x)=t^pf_i(\vec x)$.
This latter fact gives us that $\large\Sigma_{j=1}^nx_j{\partial f_i\over \partial x_j}(\vec x)=pf_i(\vec x)$.
I don't know whether it's all the summations involved or what, but I can't seem to get this to work out.  It so happens that I always have the j's and i's mixed in the wrong way...This might indicate that the question isn't true as stated...or more likely that I'm failing at some fairly basic bookkeeping.
In the end I keep ending up with $\large dg={1\over{p+1}}\Sigma^n_{j=1}\Sigma^n_{i=1}x_i{\partial f_i\over \partial x_j}dx_j+f_i$, which won't really allow me to use any of the nice identities I've earned.
 A: Let $$\alpha = \sum_{i = 1}^{n} f_{i}\ \textrm{d}x_{i}$$
be a closed one-form with functions $f_{i}$ smooth and homogeneous of degree $p$.
Since $\alpha$ is closed, $\textrm{d}\alpha = 0$ and so
$$\sum_{j = 1}^{n}\sum_{i = 1}^{n}\frac{\partial f_{i}}{\partial x_{j}} \textrm{d}x_{j}\wedge \textrm{d}x_{i} = 0, $$
$$\Rightarrow \frac{\partial f_{i}}{\partial x_{j}} = \frac{\partial f_{j}}{\partial x_{i}}.$$
Let $g$ be defined by
$$g = \frac{1}{p+1} \sum_{i = 1}^{n}x_{i}f_{i}.$$
Then
$$\textrm{d}g = \frac{1}{p+1}\left(\sum_{j=1}^{n}f_{j}\ \textrm{d}x_{j} + \left(\sum_{i=1}^{n} x_{i}\frac{\partial f_{i}}{\partial x_{j}} \right)\textrm{d}x_{j}\right).$$
Use the relationship we derived from the condition that $\alpha$ is closed to swap the indices inside the innermost sum:
$$\textrm{d}g = \frac{1}{p+1}\left(\sum_{j=1}^{n}f_{j}\ \textrm{d}x_{j} + \left(\sum_{i=1}^{n} x_{i}\frac{\partial f_{j}}{\partial x_{i}} \right)\textrm{d}x_{j}\right).$$
Now use the identity you derived in your question (after "The latter fact...")
$$\textrm{d}g = \frac{1}{p+1}\left(\sum_{j=1}^{n}f_{j}\ \textrm{d}x_{j} + p f_{j}\ \textrm{d}x_{j}\right),$$
$$\textrm{d}g = \sum_{j = 1}^{n} f_{j}\ \textrm{d}x_{j}$$
as required.
