Determinant and alternating multilinear function Let $V$ be a vector space of dimension $n$ with basis $\{v_1,\cdots,v_n\}$.
Let $\phi$ be an n-alternating multilinear map and $A:V\rightarrow V$ is any map (matrix form)  then we have to prove  that 
$\phi(Av_1,\cdots,Av_n)=\det(A) \phi(v_1,v_2,\cdots,v_n)$
I could do this for $n=2$ but could not generalize it..
$$\phi(Av_1,Av_2)=\phi(a_{11}v_1+a_{12}v_2,a_{21}v_1+a_{22}v_2)$$
$$=a_{11}a_{21}\phi(v_1,v_1)+a_{11}a_{22}\phi(v_1,v_2)+a_{12}a_{21}\phi(v_2,v_1)+a_{12}a_{22}\phi(v_2,v_2)$$
As $\phi$ is alternating $\phi(v_1,v_1)=\phi(v_2,v_2)=0$ and $\phi(v_2,v_1)=-\phi(v_1,v_2)$. 
So, we have 
$\phi(Av_1,Av_2)=a_{11}a_{22}\phi(v_1,v_2)-a_{12}a_{21}\phi(v_1,v_2)=(a_{11}a_{22}-a_{12}a_{21})\phi(v_1,v_2)=\det(A)\phi(v_1,v_2)$
Now, i am having difficulty in generalizing this..
$$\det(A)\phi(v_1,v_2,\cdots,v_n)=\sum_{\sigma}sgn(\sigma)\Pi_{i} a_{i\sigma(i)}\phi(v_1,v_2,\cdots,v_n)$$
As $\phi$ is alternating we have $sgn(\sigma)\phi(v_1,v_2,\cdots,v_n)=\phi(v_{\sigma(1)},v_{\sigma(2)},\cdots,v_{\sigma(n)})$
So, we have 
$$\det(A)\phi(v_1,v_2,\cdots,v_n)=\sum_{\sigma}\Pi_{i} a_{i\sigma(i)}\phi(v_{\sigma(1)},v_{\sigma(2)},\cdots,v_{\sigma(n)})$$
I am not able to relate this to $\phi(Av_1,\cdots,Av_n)$.
See that $A=(a_{ij})$..
Please provide some hints
 A: The map $V^n\to K$,
$(v_1,\ldots, v_n)\mapsto \phi(Av_1,\ldots, Av_n)$ is readily checked to be alternating multilinear. As the space of alternating $n$-forms is onedimensional, it must be a multiple $c(A)\phi$ of $\phi$. To determine $c(A)$ it suffices to evaluate $\phi(e_1,\ldots, e_n)$, i.e., $\phi$ applied to the columns of $A$. By alternating multilinear property, adding a column to another column does not change the value of $\phi$, whereas multiplying a column with a constant multiplies the value by the same constant. As these (Gauß) steps can be used to transform $A$ to the identity matrix while extracting $\det(A)$ as factor, we conclude that indeed $c(A)=\det(A)$.
A: It seems as though you have the right idea, but you're getting caught up on the notation.  Here's a suitable generalization of your proof:
$$
\begin{align}
\phi(Av_1,\cdots,Av_n) &= 
\phi\left(\sum_{j_1=1}^n a_{1j_1} v_{j_1}, \sum_{j_2=1}^n a_{2j_2} v_{j_2}, \dots, \sum_{j_n=1}^n a_{nj_n} v_{j_n}\right)
\\ & =
\sum_{j_1=1}^n \sum_{j_2=1}^n \cdots \sum_{j_n=1}^n
\left(\left[\prod_{i=1}^n a_{i j_i}\right] \phi\left( v_{j_1}, v_{j_2}, \dots, v_{j_n}\right)\right)
\\ & =
\sum_{(j_1,\dots,j_n) \in \{1,\dots,n\}^n}
\left(\left[\prod_{i=1}^n a_{i j_i}\right] \phi\left( v_{j_1}, v_{j_2}, \dots, v_{j_n}\right)\right)
\end{align}
$$
Now, we note that if any two of the $j_{k}$ are equal, then $\phi\left( v_{j_1}, v_{j_2}, \dots, v_{j_n}\right) = 0$, so that we may remove the corresponding term from the sum.  We may simply sum over all tuples $(j_1,\dots,j_n)$ that (may) produce a non-zero term, which means we want all tuples of $j_k$ such that each pair of $j_k$ are mutually distinct.  That is, $(j_1,\dots,j_n)$ must be a permutation of $(1,\dots,n)$.  That is, the only tuples we want to consider are those for which there is a $\sigma \in S_n$ such that $j_k = \sigma(k)$. We may therefore rewrite the above sum as
\begin{align}
\phi(Av_1,\cdots,Av_n) &= 
\sum_{\sigma \in S_n}
\left(\left[\prod_{i=1}^n a_{i \sigma(i)}\right] \phi\left( v_{\sigma(1)}, v_{\sigma(2)}, \dots, v_{\sigma(n)}\right)\right)
\\ & = 
\sum_{\sigma \in S_n}
\left(\left[\prod_{i=1}^n a_{i \sigma(i)}\right] \operatorname{sgn}(\sigma)\, \phi\left( v_{1}, v_{2}, \dots, v_{n}\right)\right)
\\ & = 
\left(\sum_{\sigma \in S_n}
\left[\prod_{i=1}^n a_{i \sigma(i)}\right] \operatorname{sgn}(\sigma)\right) \phi\left( v_{1}, v_{2}, \dots, v_{n}\right)
\end{align}
