Doubt in the proof of Cauchy-Binet formula. I have a doubt in the proof of the Cauchy-Binet formula:

Question:
In the colored portion.
I dont know how the sum multiplying $det(A(J'))$ become 
$$\sum_\sigma sgn(\sigma) b_{j'_{\tau(1)}1} b_{j'_{\tau(2)}2}\ldots  b_{j'_{\tau(k)}k}$$
Help me in that step.
Thanks in advance.
 A: The key step that your probably missing is this one. The sum 
$$\sum_{j_1,j_2,...,j_k=1}^n\det(A(j_1,j_2,...,j_k))b_{j_1}b_{j_2}...b_{j_k}$$
is taken over all multi-indices, not just the ordered ones $j_1<j_2...<j_k$. So what the author is doing is decomposing the set of all muti-indices into those which are of the form $j_1<j_2...<j_k$ and the permutations of these. So we will fix any ordered multi-index in $J^{\prime}$, then sum over the permutations of this index and lastly sum the results over $J^{\prime}$. 
So fix a multi-index $j^{\prime}_1<...<j^{\prime}_k$, and let $\sigma$ index the permutations of this index. We have
$$\sum_{\sigma}\det(A(j_1,j_2,...,j_k))b_{j_1}b_{j_2}...b_{j_k}=\sum_{\sigma}sgn(\sigma)\det(A(j_1,j_2,...,j_k))sgn(\sigma)b_{j_1}b_{j_2}...b_{j_k}$$
Using the fact that $$sgn(\sigma)\det(A(j_1,j_2,...,j_k))=\det(A(j^{\prime}_1,j^{\prime}_2,...,j^{\prime}_k))$$
you are left with 
$$\sum_{\sigma}sgn(\sigma)\det(A(j_1,j_2,...,j_k))sgn(\sigma)b_{j_1}b_{j_2}...b_{j_k}=\det(A(j^{\prime}_1,j^{\prime}_2,...,j^{\prime}_k))\sum_{\sigma}sgn(\sigma)b_{j_1}b_{j_2}...b_{j_k}$$
Then use that $j_{i}=j^{\prime}_{\tau(i)}$
and your done.
By the way, if its any consolation, I think that this proof was written in a terribly confusing way myself. 
