Let $V$ be a vector space of dimension $2n$, $T$ be an inner product on $V$, $v_1, v_2, \cdots, v_{2n}$ be an orthonormal basis and $v_1^*, v_2^*, \cdots v_{2n}^*$ be the dual basis. Let $$\omega = v_1^* \wedge v_2^* + v_3^* \wedge v_4^* + \cdots + v_{2n-1}^* \wedge v_{2n}^*$$ be an alternating 2-tensor. Prove that $$\frac{1}{n!}\omega^n = \frac{1}{n!}\omega \wedge \omega \wedge \cdots \wedge \omega$$ is a volume element of V, where there are $n$ copies of $\omega$ in the product.
Also prove that there do not exist linear functions $\gamma_1, \cdots , \gamma_{2n-2} \in V^*$ such that $$\omega = \gamma_1 \wedge \gamma_2 + \gamma_3 \wedge \gamma_4 + \cdots + \gamma_{2n-3} \wedge \gamma_{2n-2}.$$
Definition of a Volume Element from Spivak: If an orientation $\mu$ for $V$ has also been given, it follows that there is a unique $\omega$ $\in \wedge^n V$ such that $w(v_1, ... ,v_n) = 1$ whenever $v_1, ... ,v_n$ is an orthonormal basis such that $[v_1, ... ,v_n] = \mu$. This unique $\omega$ is called the volume element of $V$, determined by the inner product $T$ and orientation $\mu$.
I am given no orientation for $V$, so how might I go about finding one or is it even necessary given what is offered in the question. Also, the question implies that there are many volume elements of V, yet the definition says it is unique.