Putting the wedge product in standard/normal form I have to compute the wedge product of
$$(e_1^* + ze_2^*) \wedge (e_2^* + ze_3^*) \wedge \cdots \wedge (e_{n-1}^* + ze_n^*) \wedge (e_n^* + ze_1^*),$$
and then put it in normal/standard form. 
So I did this and got 
$$(e_1^* \wedge e_n^* \wedge e_{n-1}^* \wedge \cdots \wedge e_2^*) + z^n (e_1^* \wedge e_2^* \wedge e_{3}^* \wedge \cdots \wedge e_1^*).$$
Now I thought both weren't in standard form as the second one needs the $e_1^*$ to move to the front and if I moved $e_1^*$ to the back in the first one, that would put it in "reverse" standard form. But apparently, the first bit is in standard form. Why is this?
 A: Let us put $w=e_1^* \wedge e_2^* \wedge e_{3}^* \wedge \cdots \wedge e_n^*$. 
Then  for any permutation $\sigma$ of the indices $1,2, \ldots ,n$, we have
$$
e_{\sigma(1)}^* \wedge e_{\sigma(2)}^* \wedge e_{\sigma(3)}^* \wedge \cdots 
\wedge e_{\sigma(n)}^* = \varepsilon(\sigma) w
$$
where $\varepsilon(\sigma)$ is the signature of the permutation $\sigma$.
In your case, what you have is the permutation $\alpha_n$ defined by
$$
\alpha_n(1)=1, \alpha_n(2)=n, \alpha_n(3)=n-1, \ldots, \alpha_n(n)=2 
$$
and the permutation $\beta_n$ defined by
$$
\beta_n(1)=2, \beta_n(2)=3, \beta_n(3)=4, \ldots, \beta_n(n-1)=n, , \beta_n(n)=1 
$$
Now $\beta_n$ is a cyle of length $n$, so $\varepsilon(\beta_n)=(-1)^{(n-1)}$. Also,
$\alpha_n$ is the product of the transpositions $(2,n),(3,n-1), \ldots$ etc.
To be precise : if $n$ is even, say $n=2q$, then $\alpha_n$ is the product of
all the transpositions $(j+1,n+2-j)$ for $1 \leq j \leq q$, so $\varepsilon(\alpha_n)=
(-1)^q$. If $n$ is odd, say $n=2q+1$, then $\alpha_n$ is the product of
all the transpositions $(j+1,n+2-j)$ for $1 \leq j \leq q$, so $\varepsilon(\alpha_n)=
(-1)^q$.
To summarize, for all $n$ we have $\varepsilon(\alpha_n)=(-1)^{\lfloor\frac{n}{2} \rfloor}$.
so
$\varepsilon(\alpha_n)=-1$. So if one denotes by $E$ the expression you want to compute,
$$
\begin{array}{lcl}
E &=& (e_1^* \wedge e_n^* \wedge e_{n-1}^* \wedge \cdots \wedge e_2^*) + z^n (e_2^* \wedge e_{3}^* \wedge \cdots \wedge e_1^*) \\
&=& e_{\alpha(1)}^* \wedge e_{\alpha(2)}^* \wedge e_{\alpha(3)}^* \wedge \cdots 
\wedge e_{\alpha(n)}^* + z^n(e_{\beta(1)}^* \wedge e_{\beta(2)}^* \wedge e_{\beta(3)}^* \wedge \cdots 
\wedge e_{\beta(n)}^*) \\
&=& \varepsilon(\alpha_n)w+z^n\varepsilon(\beta_n)w \\
&=& \bigg( (-1)^{\lfloor\frac{n}{2} \rfloor}-(-z)^n \bigg) e_1^* \wedge e_2^* \wedge e_{3}^* \wedge \cdots \wedge e_n^*
\end{array}
$$
