Compute the value of the exterior $2$-form 
Compute the value of the exterior $2$-form
$$\omega = (x_1 + x_2)e_1^* \wedge e_2^* + (x_2 + x_3)e_2^* \wedge e_3^* + \cdots + (x_i, x_{i+1})e_i^* \wedge e_{i+1}^*+\cdots+(x_{n-1},x_n)e_{n-1}^* \wedge e_n^*$$
on the pair of vectors $(x_1e_1 + \cdots + x_ne_n, e_1 + \cdots + e_n)$, where $x_1, \ldots, x_n$ are variables.

I think it has to be in the form
$$\frac{1}{n!} \det(A) $$
where $A$ is the matrix you get from applying each $e_i^*$ to each $e_i$, but I'm having trouble working out this matrix because of the way $\omega$ is. I thought about expanding the brackets and then applying the vectors to each other to get my $n \times n$ matrix. Is this correct?
So expanding the brackets will give me
$$(x_1e_1^*) \wedge e_2^* + (x_2e_1^*)e_2^* + \cdots + (x_ie_i^*)\wedge e_{i+1}^* + (x_{i+1}e_i^*) \wedge e_{i+1}^* + \cdots + (x_{n-1}e_{n-1}*) \wedge e_{n} + (x_ne_{n-1}^*) \wedge e_n^*$$
Once I've expanded the brackets, can I say $(x_1e_1^*) \wedge e_2^* = x_1e_1^* \wedge e_2^*$, which when I then apply to the vectors I would get $x_1 + 1$ as in the first vector we have $x_1e_1$ and so then I only get the coeffeciant when $x_1e_1^*$ acts on it, similary with the second vector and when $e_2^*$ acts on it?
 A: You might consider forming the 2-vector $\sum_{i,j} (x_i e_i) \wedge e_j$.  You can then consider the action of the 2-form on the 2-vector just by matching up similar components:  the component of $e_1^* \wedge e_2^*$ multiplies the component of $e_1 \wedge e_2$, and so on.  Only these corresponding components multiply and contribute to the scalar result.
In essence, this is no different than how you would approach the scalar resultant of a 1-form and a (1-)vector--by multiplying corresponding components and adding up those products.
Having the 2-form act on one of the vectors to yield a 1-form and then having that 1-form act on the remaining vector is perfectly valid but, in my opinion, tedious by comparison.  Evaluating the 2-vector first allows you to disregard a lot of terms that you know won't contribute to the end result.
Here's a sample of how this computation would go.  Let there be a 2-form $\omega = (\alpha + \beta) e_{12}^* + (\beta + \gamma) e_{23}^*$.  Let $u = \alpha e_1 + \beta e_2 + \gamma e_3$ and $v = e_1 + e_2 + e_3$.
Calculate $u \wedge v = (\alpha - \beta) e_{12} + (\beta - \gamma) e_{23} + (\gamma - \alpha) e_{31}$.  When we find the scalar product of $\omega$ and $u \wedge v$, we know that the $e_{31}$ term cannot contribute because $\omega$ has no $e^*_{31}$ component.  You can use this property in advance to avoid calculating this product at all when you evaluate $u \wedge v$, saving yourself some work.
I'm not entirely sure what your conventions are, but when you find the scalar product of $\omega$ and $u \wedge v$, you should get $\alpha^2 - \beta^2 + \beta^2 - \gamma^2 = \alpha^2 - \gamma^2$ within, perhaps, an overall factor of a minus sign.
A: Use the relations $(e_i^*\wedge e_j^*)(e_k,e_l)=\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}$ via linearity: 
$$\left[\sum_{k=1}^{n-1}(x_k+x_{k+1})(e_k^*\wedge e_{k+1}^*)\right]\left(\sum_{j=1}^n x_je_j,\sum_{l=1}^n e_l\right)=\sum_{k,j,l}(x_k+x_{k+1})x_j(e_k^*\wedge e_{k+1}^*)(e_j,e_l)$$
Can you finish from here?
