Sign of the $n$-form

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Consider $\alpha_1,\dots,\alpha_n$ 1-form in a manifold. Take $\omega = \alpha_1 \wedge \cdots \wedge \alpha_n$. Now, fixed $i,j \in \{1,\dots,n\}$ and $\sigma$ a permutation such that $\sigma(k) = k$ for $k \neq i,j$ and $\sigma(i) = j$ and $\sigma(j) = i$, is this formula correct? $$ \alpha_{\sigma(1)}\wedge \cdots \wedge \alpha_{\sigma(n)} = (-1)^{i+j} \alpha_1 \wedge \cdots\wedge\alpha_n $$

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Counterexample: $dx^3\wedge dx^2 \wedge dx^1 = - dx^1\wedge dx^2 \wedge dx^3 \neq (-1)^{1+3} dx^1\wedge dx^2\wedge dx^3$ in $\Bbb R^3$. What is true is that for $\sigma\in \mathfrak{S}_n$, $\alpha^{\sigma(1)}\wedge \cdots \wedge \alpha^{\sigma(n)} = \varepsilon(\sigma) \alpha^1 \wedge \cdots \wedge \alpha^n$, where $\varepsilon(\sigma)$ is the signature of $\sigma$. In your specific question, $\sigma$ is just the transposition $(ij)$, which has signature $-1$.