- Suppose that $\phi_1,\dots,\phi_k\in V^\ast$ and $v_1,\dots,v_k\in V$ where $\dim V=k$. Prove that $$\phi_1\wedge\dots\wedge\phi_k(v_1,\dots,v_k)=\frac{1}{k!}\det[\phi_i(v_j)].$$
- More generally, show that whenever $\phi_1,\dots,\phi_p\in V^\ast$ and and $v_1,\dots,v_p\in V$, $$\phi_1\wedge\dots\wedge\phi_p(v_1,\dots,v_p)=\frac{1}{p!}\det[\phi_i(v_j)].$$
Remark about definitions: If $T$ is a $p$-tensor on $V$, then $$\operatorname{Alt}(T)(v_1,\dots,v_p)=\frac{1}{p!}\sum_{\sigma \in S_p}\operatorname{sgn}\sigma\cdot T(v_{\sigma(1)},\dots,v_{\sigma(n)})$$ and if $T_i\in \Lambda^i(V^\ast)$ for $i=1,\dots, n$, then $$T_1\wedge\dots\wedge T_n=\operatorname{Alt}(T_1\otimes\dots\otimes T_n).$$
I can prove the first part: $$\phi_1\wedge\dots\wedge \phi_k(v_1,\dots,v_k)=\operatorname{Alt}(\phi_1\otimes\dots\otimes \phi_k)(v_1,\dots,v_k)=\\\frac{1}{k!}\sum_{\sigma\in S_k}\operatorname{sgn}\sigma \cdot \phi_1(v_{\sigma(1)})\dots\phi_k(v_{\sigma(k)})=\frac{1}{k!}\det[\phi_i(v_j)].$$
But how do I show the second part?