设 $\varphi_i\in\Lambda^1(V)$, $i=1,2,\ldots,k$. 证明: $\varphi_1\wedge\cdots\wedge\varphi_k(v_1,\ldots,v_k)=\det(\varphi_i(v_j))$.
设 $\varphi_i\in\Lambda^1(V)$, $i=1,2,\ldots,k$. 证明:
\[\varphi_1\wedge\cdots\wedge\varphi_k(v_1,\ldots,v_k)=\det(\varphi_i(v_j)).\]