Topology. Matrix question

Let $\displaystyle x_1 , x_2, ..., x_k$ be vectors in $\displaystyle \mathbb{R}^n$; let $\displaystyle X$ be a matrix $\displaystyle X = [x_1 , x_2, ..., x_k]$. If $\displaystyle I = (i_1, i_2,...,i_k)$ is an arbitrary $\displaystyle k$-tuple from the set $\displaystyle \{1, 2, ... ,n\}$, show that

$\displaystyle \phi_{i_1} \wedge \phi_{i_2} \wedge ... \wedge \phi_{i_k} (x_1 , x_2, ..., x_k) = det X_I$,

where

$\displaystyle \phi_{i}(e_j) = \begin{cases} 1, & \mbox{if } i=j \\ 0, & \mbox{if } i \neq j \end{cases}$

for $\displaystyle i, j \in \{1,2,...,n\}.$

Please help or give a hint

Re: Topology. Matrix question

write out the definition of the wedge product. then write out your vectors in the basis ej. Then using what you have above, and the fact that each phi i is linear, you realise that the only non zero component in your sum is when each phi i is applied to just ei. ie. phi1(e1).phi2(e2)...but this is just the definition of the determinant of the matrix X