[proofplan] We prove both directions. The forward direction ($\omega$ non-degenerate $\Rightarrow \omega^n \neq 0$) uses the [Standard Form Theorem](/theorems/1548) to choose a basis of $V$ in which $\omega = \sum_{j=1}^n e_{2j-1} \wedge e_{2j}$ (viewing $\omega \in \Lambda^2(V^*)$ in the dual basis). A direct multinomial expansion of $\omega^n = \omega \wedge \cdots \wedge \omega$ collapses to a single non-zero top-degree form $n!\, e_1 \wedge e_2 \wedge \cdots \wedge e_{2n}$, so $\omega^n \neq 0$. For the reverse direction we argue by contrapositive: if $\omega$ is degenerate, there is a non-zero $v \in V$ with $\omega(v, \cdot) = 0$, i.e. the contraction $\iota_v \omega = 0$. The derivation property of $\iota_v$ yields $\iota_v(\omega^n) = n\, (\iota_v \omega) \wedge \omega^{n-1} = 0$; and $\iota_v$ annihilates a non-zero top-form if and only if that top-form is itself zero (since contraction by a non-zero vector is surjective in top degree). Hence $\omega^n = 0$. [/proofplan]

[step:Reduce to the standard basis via the Standard Form Theorem] Suppose $\omega$ is non-degenerate. By the [Standard Form Theorem](/theorems/1548), whose hypotheses ($V$ finite-dimensional real, $\omega$ non-degenerate skew-symmetric bilinear) are exactly our hypotheses, there exists a basis $e_1, e_2, \dots, e_{2n}$ of $V^*$ such that \begin{align*} \omega = \sum_{j=1}^n e_{2j-1} \wedge e_{2j} \in \Lambda^2(V^*), \end{align*} where we identify the bilinear form $\omega$ with its associated $2$-form via $\omega(u, v) = \omega(u, v)$ (anti-symmetrisation gives the same object up to convention; we fix the convention so that in the dual basis of $\{f_i\}$ to $\{e_i\}$, $\omega(f_{2j-1}, f_{2j}) = 1$). In particular, $\omega$ is expressed as a sum of $n$ pairwise "independent" wedge monomials: each $e_{2j-1} \wedge e_{2j}$ involves only the two basis elements $e_{2j-1}, e_{2j}$, and these pairs are disjoint for different $j$. [/step]

[step:Forward direction: compute $\omega^n$ by multinomial expansion]Let $\alpha_j := e_{2j-1} \wedge e_{2j} \in \Lambda^2(V^*)$ for $j = 1, \dots, n$. Then $\omega = \sum_{j=1}^n \alpha_j$, and \begin{align*} \omega^n = \bigg( \sum_{j=1}^n \alpha_j \bigg)^n = \sum_{(j_1, \dots, j_n) \in \{1, \dots, n\}^n} \alpha_{j_1} \wedge \alpha_{j_2} \wedge \cdots \wedge \alpha_{j_n}. \end{align*} We use: **(a) $\alpha_j \wedge \alpha_k = \alpha_k \wedge \alpha_j$ for all $j, k$.** Each $\alpha_j$ is a $2$-form, so the Koszul sign for transposition is $(-1)^{2 \cdot 2} = 1$. **(b) $\alpha_j \wedge \alpha_j = 0$ for all $j$.** This is because $\alpha_j = e_{2j-1} \wedge e_{2j}$ involves only two $1$-form factors, and squaring gives $e_{2j-1} \wedge e_{2j} \wedge e_{2j-1} \wedge e_{2j}$, which contains a repeated factor $e_{2j-1}$ (or equivalently $e_{2j}$), hence vanishes. By (a), reorder each term so that the indices appear in non-decreasing order. By (b), any term with a repeated index vanishes. The only surviving multi-indices are permutations of $(1, 2, \dots, n)$, and by (a) all such permutations give the same product $\alpha_1 \wedge \alpha_2 \wedge \cdots \wedge \alpha_n$. The number of permutations of $(1, \dots, n)$ is $n!$, so \begin{align*} \omega^n = n!\, \alpha_1 \wedge \alpha_2 \wedge \cdots \wedge \alpha_n = n!\, (e_1 \wedge e_2) \wedge (e_3 \wedge e_4) \wedge \cdots \wedge (e_{2n-1} \wedge e_{2n}) = n!\, e_1 \wedge e_2 \wedge \cdots \wedge e_{2n}. \end{align*} The form $e_1 \wedge e_2 \wedge \cdots \wedge e_{2n}$ is the wedge product of all $2n$ basis $1$-forms, which is a non-zero element of the $1$-dimensional space $\Lambda^{2n}(V^*)$. Since $n! > 0$, we conclude $\omega^n \neq 0$.[/step]

[guided]We want to show the $n$-th wedge power of $\omega$ does not vanish. The strategy is to use the standard-form basis — in that basis $\omega$ is a sum of $n$ "commuting" building blocks $\alpha_j = e_{2j-1} \wedge e_{2j}$, each of which squares to zero but together produce the top form. **Step A: Multinomial expansion.** By multilinearity of the wedge product, \begin{align*} \omega^n = \bigg(\sum_{j=1}^n \alpha_j\bigg)^n = \sum_{(j_1, \dots, j_n)} \alpha_{j_1} \wedge \cdots \wedge \alpha_{j_n}. \end{align*} **Step B: Commutativity of the $\alpha_j$.** Each $\alpha_j$ is a $2$-form (even-degree), and the Koszul sign for swapping two even-degree forms is $(-1)^{2 \cdot 2} = +1$. So the $\alpha_j$ pairwise commute: $\alpha_j \wedge \alpha_k = \alpha_k \wedge \alpha_j$. **Step C: $\alpha_j^2 = 0$.** Writing $\alpha_j \wedge \alpha_j = e_{2j-1} \wedge e_{2j} \wedge e_{2j-1} \wedge e_{2j}$, we see a repeated factor (either $e_{2j-1}$ or $e_{2j}$), and any wedge of $1$-forms with a repeated factor is zero. **Step D: Collapse the sum.** Using (B), we may reorder indices; using (C), any term with a repeated index vanishes. Surviving multi-indices are exactly permutations of $(1, 2, \dots, n)$. All such permutations produce the same product $\alpha_1 \wedge \cdots \wedge \alpha_n$ by (B), and there are $n!$ of them: \begin{align*} \omega^n = n!\, \alpha_1 \wedge \alpha_2 \wedge \cdots \wedge \alpha_n. \end{align*} **Step E: Rewrite in the $e_i$ basis.** Each $\alpha_j = e_{2j-1} \wedge e_{2j}$, and concatenation gives \begin{align*} \alpha_1 \wedge \cdots \wedge \alpha_n = e_1 \wedge e_2 \wedge e_3 \wedge e_4 \wedge \cdots \wedge e_{2n-1} \wedge e_{2n}, \end{align*} which is a non-zero element of the $1$-dimensional top exterior power $\Lambda^{2n}(V^*)$. Multiplying by $n! > 0$ preserves non-zeroness, so $\omega^n \neq 0$.[/guided]

[step:Reverse direction: set up the contrapositive and use the contraction $\iota_v$] We prove: if $\omega$ is degenerate, then $\omega^n = 0$. This is the contrapositive of "$\omega^n \neq 0 \Rightarrow \omega$ non-degenerate", which together with the forward direction proves the claimed equivalence. Suppose $\omega$ is degenerate. By definition of degeneracy, there exists $v \in V$ with $v \neq 0$ such that \begin{align*} \omega(v, w) = 0 \qquad \text{for all } w \in V. \end{align*} In the language of interior product (contraction) on the exterior algebra: the interior product $\iota_v: \Lambda^k(V^*) \to \Lambda^{k-1}(V^*)$ is defined on $1$-forms by $\iota_v \eta = \eta(v)$ and extended to the full exterior algebra by the graded derivation property. For a $2$-form $\omega \in \Lambda^2(V^*)$: \begin{align*} \iota_v \omega \in \Lambda^1(V^*), \qquad (\iota_v \omega)(w) = \omega(v, w) \text{ for all } w \in V. \end{align*} By hypothesis $\omega(v, w) = 0$ for all $w$, so \begin{align*} \iota_v \omega = 0. \end{align*} [/step]

[step:Apply the graded derivation property to conclude $\iota_v \omega^n = 0$]The interior product $\iota_v$ is a graded derivation of degree $-1$ on the exterior algebra $\Lambda^\bullet(V^*)$: for $\eta \in \Lambda^p(V^*)$ and $\zeta \in \Lambda^q(V^*)$, \begin{align*} \iota_v (\eta \wedge \zeta) = (\iota_v \eta) \wedge \zeta + (-1)^p\, \eta \wedge (\iota_v \zeta). \end{align*} Apply this inductively to $\omega^n = \omega \wedge \omega^{n-1}$: since $\omega$ is a $2$-form (so $p = 2$, $(-1)^p = 1$), \begin{align*} \iota_v \omega^n = (\iota_v \omega) \wedge \omega^{n-1} + \omega \wedge \iota_v(\omega^{n-1}). \end{align*} The first term vanishes because $\iota_v \omega = 0$. By induction on $n \geq 1$ (with base case $\iota_v \omega = 0$), the second term also vanishes: assuming $\iota_v \omega^{n-1} = 0$, we get $\omega \wedge \iota_v(\omega^{n-1}) = 0$. The induction step from $n-1$ to $n$ is exactly the computation above, with both terms zero. Therefore for all $n \geq 1$, \begin{align*} \iota_v \omega^n = 0. \end{align*} Equivalently, by the derivation formula applied all the way out, $\iota_v \omega^n = n\, (\iota_v \omega) \wedge \omega^{n-1} = 0$.[/step]

[guided]We want to exploit degeneracy of $\omega$ — equivalently, $\iota_v \omega = 0$ for some $v \neq 0$ — to drive a chain reaction: $\iota_v \omega^n = 0$, and then show this forces $\omega^n$ itself to vanish. **The derivation formula.** For any graded derivation $D$ of degree $d$ on an algebra and elements $\eta, \zeta$ of degrees $p, q$: \begin{align*} D(\eta \wedge \zeta) = D\eta \wedge \zeta + (-1)^{dp}\, \eta \wedge D\zeta. \end{align*} For $D = \iota_v$ (degree $-1$) and $\eta = \omega$ (degree $2$), the Koszul sign is $(-1)^{-1 \cdot 2} = +1$. So \begin{align*} \iota_v(\omega \wedge \omega^{n-1}) = (\iota_v \omega) \wedge \omega^{n-1} + \omega \wedge \iota_v(\omega^{n-1}). \end{align*} **Induction.** By inductive hypothesis $\iota_v \omega^{n-1} = 0$ (with base $n-1 = 0$: $\iota_v(1) = 0$), both terms on the right vanish, giving $\iota_v \omega^n = 0$. **Alternative closed-form derivation.** By applying the Leibniz rule to each of the $n$ factors of $\omega^n = \omega \wedge \omega \wedge \cdots \wedge \omega$, we see $\iota_v$ is distributed over all $n$ positions; symmetry and the commutativity of $2$-forms collect them: \begin{align*} \iota_v(\omega^n) = n\, (\iota_v \omega) \wedge \omega^{n-1} = n \cdot 0 \cdot \omega^{n-1} = 0. \end{align*}[/guided]

[step:Conclude $\omega^n = 0$ from $\iota_v \omega^n = 0$ via non-degeneracy of contraction on top forms] We claim: if $v \in V$ is non-zero and $\mu \in \Lambda^{2n}(V^*)$ is a top-degree form with $\iota_v \mu = 0$, then $\mu = 0$. **Proof of the claim.** Extend $v$ to a basis $v = f_1, f_2, \dots, f_{2n}$ of $V$, and let $e_1, \dots, e_{2n}$ be the dual basis of $V^*$. The space $\Lambda^{2n}(V^*)$ is $1$-dimensional, spanned by $e_1 \wedge \cdots \wedge e_{2n}$, so every $\mu \in \Lambda^{2n}(V^*)$ is of the form $\mu = c\, e_1 \wedge \cdots \wedge e_{2n}$ for some $c \in \mathbb{R}$. Compute \begin{align*} \iota_v \mu = \iota_{f_1}(c\, e_1 \wedge e_2 \wedge \cdots \wedge e_{2n}) = c\, e_1(f_1)\, e_2 \wedge \cdots \wedge e_{2n} = c\, e_2 \wedge e_3 \wedge \cdots \wedge e_{2n}, \end{align*} using the derivation rule and $e_i(f_1) = \delta_{i, 1}$. The form $e_2 \wedge e_3 \wedge \cdots \wedge e_{2n} \in \Lambda^{2n-1}(V^*)$ is non-zero (a basis element of the top-minus-one exterior power). Therefore $\iota_v \mu = 0$ forces $c = 0$, i.e. $\mu = 0$. **Apply the claim with $\mu = \omega^n$.** By the previous step, $\iota_v \omega^n = 0$ with $v \neq 0$. By the claim, $\omega^n = 0$. This completes the reverse direction (in its contrapositive form): degeneracy of $\omega$ implies $\omega^n = 0$. Combined with the forward direction, we have \begin{align*} \omega \text{ non-degenerate} \iff \omega^n \neq 0, \end{align*} which is the statement of the theorem. [/step]