I have proved these two exercises:
(1) Suppose that $T \in \Lambda^p(V^*)$ and $v_1, \ldots, v_p \in V$ are linearly dependent. Prove that $T(v_1, \ldots, v_p) = 0$ for all $T \in \Lambda^p(V^*)$.
And
(2) Suppose that $\phi_1, \ldots, \phi_k \in V^*$, and $v_1, \ldots, v_k \in V$, where $k$ = dim$V$. Prove that $$\phi_1 \wedge \cdots \wedge \phi_k (v_1, \cdots, v_k) = \frac{1}{k!}\det[\phi_i(v_j)].$$
But now I am asked to prove
Show that whenever $\phi_1, \ldots, \phi_k \in V^*$, and $v_1, \ldots, v_k \in V$, then $$\phi_1 \wedge \cdots \wedge \phi_p (v_1, \cdots, v_p) = \frac{1}{k!}\det[\phi_i(v_j)].$$
But my concern is that, if $\phi_1, \ldots, \phi_k \in V^*$ are linearly dependent, it follows directly from (1), otherwise, (2) can be extend to any sub-dimension. So, that's it?