Show that $d\omega=(-1)^{n-1} (divX)dx_1 \wedge\dots\wedge dx_n$

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Let, $X = (X_1,\dots,X_n): D \subset \Bbb R^n \to \Bbb R^n$ be a $C^1$ vectr field on a domain $D$. Define $divX := \sum_{i=1}^n {\frac{\partial X_i}{\partial x_i}}$. Define an $(n-1)$ form $\omega$ on $D$ by,$$\omega_p(v_1,\dots,v_{n-1}):= det[v_1|\dots|v_{n-1}|X_p] , \forall p \in D, v_1,\dots v_{n-1} \in \Bbb R^n$$ Show that $d\omega=(-1)^{n-1} (divX)dx_1 \wedge\dots\wedge dx_n$

My attempt:

I know how to compute the exterior derivative of a form when it is of the form :$$\omega = \sum_{I}{a_I dx_I}$$ i.e. to find,$$d\omega=\sum_{I}{da_I \wedge dx_I}$$. But I am unable to recognize the given form in that form.

Thanks in advance for help!