Suppose that $V$ is a vector space of dimension $2n$, and let $\omega \in \Lambda^2(V)$. Prove that the following two statements are equivalent.
(1) $\tilde{\omega} : V \rightarrow V^*$ defined by $X\mapsto \omega(X,\cdot)$ is an isomorphism.
(2) $\omega^n=\omega \wedge \omega \wedge \cdots \wedge \omega \in \Lambda^{2n}(V)$ is nonzero.
What I have tried is
Suppose that the map $\tilde{\omega}: V\rightarrow V^*$ defined by $X\rightarrow i_X \omega$ is an isomorphism. Then we have the basis $(e^1,f^1,e^2,f^2,\dots, e^n,f^n)$ of $V$ such that \begin{align*} &(1) \text{ }\omega(e^i,e^j)=\omega(f^i,f^j)=0 \hspace{3mm}\forall i,j\\ &(2)\text{ } \omega(e^i,f^j)=\begin{cases} 1 &i=j\\ 0 &i\neq j \end{cases}. \end{align*}
Then observe that
\begin{align*} \omega\wedge \omega \wedge \cdots \wedge \omega \in \Lambda^{2n}(V) \end{align*}
and that
\begin{align*} \omega\wedge \omega \wedge \cdots \wedge \omega (e^1,f^1,e^2,f^2,\dots, e^n,f^n)=\omega(e^1,f^1)\wedge \omega(e^2,f^2)\wedge \cdots \omega(e^n,f^n)=1\neq 0. \end{align*}
Therefore, $\omega\wedge \omega \wedge \cdots \wedge \omega $ is nonzero.\
Conversely, suppose that $\omega\wedge \omega \wedge \cdots \wedge \omega \in \Lambda^{2n}(V)$ is nonzero. Clearly, note that $\tilde{\omega}:V\rightarrow V^*$ defined by $\tilde{\omega}(X)=i_X(\omega)$ is a homomorphism since $i_{X+Y}\omega(Z)=\omega(X+Y,Z)=\omega(X,Z)+\omega(Y,Z)=i_X\omega(Z)+i_Y\omega(Z)$. We need to prove that $\tilde{\omega}$ is a bijection. And assume that $i_X(\omega\wedge \omega \wedge \cdots \wedge \omega)=0$ for some $X\in V\setminus \{0\}$. Then we can build a basis $\{X,r^1,r^2,\dots, r^{2n-1}\}$ of $V$. Since $\omega\wedge \omega \wedge \cdots \wedge \omega(X,r^1,r^2,\dots,r^{2n-1})=0$ and $\{X,r^1,r^2,\dots, r^{2n-1}\}$ is a basis of $V$, $\omega\wedge \omega \wedge \cdots \wedge \omega=0$ and it is a contradiction. Thus, $i_X(\omega\wedge \omega \wedge \cdots \wedge \omega)$ is nonzero for any $X\in V\setminus \{0\}$. Then note that, for given $X\in V\setminus \{0\}$, we have $Y\in V$ such that $i_X\omega(Y)=\omega(X,Y)\neq 0$ (otherwise, for any collection $\{Y^1,\dots, Y^{2n-1}\}\subseteq V$, $i_X(\omega\wedge \omega \wedge \cdots \wedge \omega)(Y^1,\dots, Y^{2n-1})=0\wedge \omega(Y^{2},Y^{3})\wedge \dots \wedge \omega(Y^{2n-2},Y^{2n-1})=0$). Then $ker \tilde{\omega}=0$, so $\tilde{\omega}$ is an injection. Then since $V$ and $V^*$ has same dimension, $\tilde{\omega}$ is a surjection so is a bijection. Therefore, $\tilde{\omega}$ is an isomorphism.
I mistakenly used the wrong computation of wedge product of alternating tensors. I am pretty sure that I can adjust a little bit from here to make solid proof. I am still thinking about but seems I am missing something. Any help would be appreciated. Thanks in advance.
I think I figure it out. I posted this solution for future.
Suppose that the map $\tilde{\omega}: V\rightarrow V^*$ defined by $X\rightarrow i_X \omega$ is an isomorphism. Then we have the basis $(e^1,f^1,e^2,f^2,\dots, e^n,f^n)$ of $V$ such that \begin{align*} &(1) \text{ }\omega(e^i,e^j)=\omega(f^i,f^j)=0 \hspace{3mm}\forall i,j\\ &(2)\text{ } \omega(e^i,f^j)=\begin{cases} 1 &i=j\\ 0 &i\neq j \end{cases}. \end{align*}
Let $\{\epsilon^i,\delta^i\}$ be a dual basis such that \begin{align*} &(1) \epsilon^i(e^j)=\begin{cases}1 &i=j \\0 &i\neq j \end{cases} &(2) \delta^i(f^j)=\begin{cases}1 &i=j \\0 &i\neq j \end{cases}\\ &(3) \epsilon^i(f^j)=0 \hspace{4mm}\forall i,j &(4) \delta^i(e^j)=0 \hspace{4mm}\forall i,j. \end{align*}
Then observe that
\begin{align*} \omega\wedge \omega \wedge \cdots \wedge \omega \in \Lambda^{2n}(V) \end{align*}
and that, since $dim(\Lambda^{2n}(V))=\bigg(\begin{matrix} 2n\\2n \end{matrix} \bigg)=1$, for some integer $\alpha\in \mathbb{R}$,
\begin{align*} \omega\wedge \omega \wedge \cdots \wedge \omega (e^1,f^1,e^2,f^2,\dots, e^n,f^n)&=\alpha(\epsilon^1\wedge \delta^1\wedge \epsilon^2\wedge \delta^2\wedge\cdots \epsilon^n\wedge \delta^n)(e^1,f^1,e^2,f^2,\dots, e^n,f^n)\\ &=\alpha\sum_{\sigma}sgn(\sigma)\epsilon^1(v_{\sigma(1)})\delta^1(v_{\sigma(2)})\cdots \epsilon^n(v_{\sigma(2n-1)})\delta^n(v_{\sigma(2n)})\\ &=\alpha(\epsilon^1(e^1)\delta^1(f^1)\cdots \epsilon^n(e^n)\delta^n(f^n))\\ &=\alpha\neq 0, \end{align*} where \begin{align*} v_i=\begin{cases} e^m &\text{ if }i=2m-1\\ f^m &\text{ if } i=2m \end{cases} \hspace{2cm} \forall i \in \mathbb{Z}\cap [1,2n]. \end{align*}
Therefore, $\omega\wedge \omega \wedge \cdots \wedge \omega $ is nonzero.\
Conversely, suppose that $\omega\wedge \omega \wedge \cdots \wedge \omega \in \Lambda^{2n}(V)$ is nonzero. Clearly, note that $\tilde{\omega}:V\rightarrow V^*$ defined by $\tilde{\omega}(X)=i_X(\omega)$ is a homomorphism since $i_{X+Y}\omega(Z)=\omega(X+Y,Z)=\omega(X,Z)+\omega(Y,Z)=i_X\omega(Z)+i_Y\omega(Z)$. We need to prove that $\tilde{\omega}$ is a bijection.\
Assume that $i_X(\omega\wedge \omega \wedge \cdots \wedge \omega)=0$ for some $X\in V\setminus \{0\}$. and $\alpha\in \mathbb{R}$ since $dim(\Lambda^{2n}(V))=1$. Then we can build a basis $\{X,r^1,r^2,\dots, r^{2n-1}\}$ of $V$. Then, $\forall v\in V$, $v$ can be represented as linear combination of the elements in basis. Then, for any $\{a^1,\dots, a^{2n}\}\subseteq V$, $\omega\wedge \omega \wedge \cdots \wedge \omega(a^1,\dots,a^{2n})=0$ since $\omega$ is multilinear map. It contradicts to our assumption. Therefore, $i_X (\omega\wedge \omega \wedge \cdots \wedge \omega)\neq 0$ $\forall X\in V\setminus \{0\}$. \
Let $X\in V\setminus \{0\}$ be given. Note that \begin{align*} i_X(\omega\wedge \omega \wedge \cdots \wedge \omega)= (i_X\omega)\wedge \omega \wedge \cdots \wedge \omega+\cdots + \omega\wedge \omega \wedge \cdots \wedge (i_X\omega)=n\{ (i_X\omega)\wedge \omega \wedge \cdots \wedge \omega \}\neq 0. \end{align*}
It implies that $(i_X\omega)\wedge \omega \wedge \cdots \wedge \omega\neq 0$, thus, we should have $(i_X\omega)\neq 0$. Thus, we have $Y\in V$ such that $(i_X\omega)(Y)=\omega(X,Y)\neq 0$. Since $X\in V\setminus \{0\}$ was arbitrary, $i_X\omega =0 \iff X=0$. Then $ker \tilde{\omega}=0$, so $\tilde{\omega}$ is an injection. Then since $V$ and $V^*$ has same dimension, $\tilde{\omega}$ is a surjection so is a bijection. Therefore, $\tilde{\omega}$ is an isomorphism.