wedge product, multilinear algebra in $\mathbb{R}^{2n}$ Denote coordinates in the space $\mathbb{R}^{2n}$ by $(x_1, y_1, \dots, x_n, y_n)$. Consider a $2$-form $$\omega = \sum_{i=1}^n x_i \wedge y_i.$$
(a) Compute$$\underbrace{\omega \wedge \dots \wedge \omega}_{n \text { times}}$$(b) Let $F: \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ be a linear map, such that $F^*\omega = \omega$. Prove that $\det F = 1$.
Progress so far: I understand for the first part, the $n$-fold wedge product with itself is given by$$(\omega)^{\wedge n} = \underbrace{\omega \wedge \dots \wedge \omega}_{n \text{ times}}  = n! (x_1 \wedge y_1) \wedge \dots \wedge (x_n \wedge y_n).$$But I need some help with approaching part (b)...
 A: b.
We have $$F^*(\omega^{\wedge n}) = (F^* \omega)^{\wedge n} \tag*{(1)}$$so $$n!F^*((x_1 \wedge y_1) \wedge \dots \wedge (x_n \wedge y_n)) = n!(x_1 \wedge y_1) \wedge \dots \wedge (x_n \wedge y_n).$$Since $(x_1 \wedge y_1) \wedge \dots \wedge (x_n \wedge y_n)$ is $($up to sign/orientation$)$ the volume form on $\mathbb{R}^{2n}$,$$F^*((x_1 \wedge y_1) \wedge \dots \wedge (x_n \wedge y_n)) = (\det F)(x_1 \wedge y_1) \wedge \dots \wedge (x_n \wedge y_n) \tag*{(2)}$$and $\det F = 1$.
Proof of $(1)$.
This follows by repeatedly applying $($i.e. with induction$)$ the identity$$F^*(\omega_1 \wedge \omega_2) = (F^*\omega_1) \wedge (F^* \omega_2),\tag*{(3)}$$valid for any $\omega_1 \in \bigwedge^k(V^*)$, $\omega_2 \in \bigwedge^l(V^*)$. To prove $(3)$, write $$F^*(\omega_1 \wedge \omega_2)(v_1, \dots, v_k, v_{k+1}, \dots, v_{k+l})$$$$=\omega_1 \wedge \omega_2\,(Fv_1, \dots, Fv_k, Fv_{k+1}, \dots, Fv_{k+1})$$$$=\sum_{i_1 < \dots < i_k,\,i_{k+1} < \dots < v_{k+l}} (-1)^{\text{inv}(i_1, \dots, i_{k+1})}\omega_1(Fv_{i_1}, \dots, Fv_{i_k})\omega_2(Fv_{i_{k+1}}, \dots, Fv_{i_{k+l}})$$$$=\sum_{i_1 < \dots < i_k,\,i_{k+1} < \dots < v_{k+l}} (-1)^{\text{inv}(i_1, \dots, i_{k+1})}F^*\omega_1(v_{i_1}, \dots, v_{i_k})F^*\omega_2(v_{i_{k+1}}, \dots, v_{i_{k+l}})$$$$=(F^*\omega_1) \wedge (F^* \omega_2)(v_1, \dots, v_k, v_{k+1}, \dots, v_{k+l}).$$
Proof of $(2)$.
Let $F: \mathbb{R}^n \to \mathbb{R}^n$. Then $F^*(x_1 \wedge \dots \wedge x_n) = cx_1 \wedge \dots \wedge x_n$ since the space of exterior $n$-forms on $\mathbb{R}^n$ is one-dimensional. To determine $c$, we check$$c = f^*(x_1 \wedge \dots \wedge x_n)(e_1, \dots, e_n) = x_1 \wedge \dots \wedge x_n\,(Fe_1, \dots, Fe_n)$$$$ = \left| Fe_1 \text{ }\dots\text{ }Fe_n\right| = |\text{Mat}_n(F)|.$$ 
