Corollary 3

Let $f : M^{n-1} \to A$ be an alternating $(n-1)$*-linear form*. If $(x_i)_{1 \leq i \leq n}$ is a sequence of n elements of M such that $x_1 \wedge x_2 \wedge \cdots \wedge x_n = 0$, then

(where we write $f(x_1, \ldots, \hat{x}_i, \ldots, x_n) = f(x_1, \ldots, x_{i-1}, x_{i+1}, \ldots, x_n)$ for $1 \leq i \leq n$.

It suffices to prove that the $n$-linear mapping

of $M^n$ to $M$ is alternating. Now, if $x_i = x_{i+1}$, all the terms in the sum on the right hand side have zero coefficients except $x_i$ and $x_{i+1}$, since $f$ is alternating; on the other hand, the coefficient of $x_i$ is

and that of $x_{i+1}$ is $(-1)^{i+1} f(x_1, \ldots, x_i, x_{i+2}, \ldots, x_n)$ and they are inverses by hypothesis.