Equivalence of the axiomatic and recursive definitions of the determinant. In Linear Algebra, Lang defines the determinant recursively by the formula: 
$$D(A) = D(A^1, \ldots, A^n) = (-1)^{i+1}a_{i1}D(A_{i1})+\cdots +(-1)^{i+n}a_{in}D(A_{in}),$$ where $A^1, \ldots,A^n$ are vectors in $K^n$ and $A_{ij}$ is the matrix created by deleting the $i$th row and $j$th column from matrix $A$. 
Lang then attempts to prove that the determinant is multilinear. 
Lang defines a function to be multilinear if for every entry $k$ of the function we have: 
$$F(A^1,\ldots ,\ B^k + C^k\ , \ldots ,\ A^n) = F(A^1,\ldots ,B^k, \ldots,A^n) + F(A^1,\ldots ,C^k,\ldots,A^n)
\\  \text{and } F(A^1\ , \ldots ,\ x\ A^k,\ldots,\ A^n)= x\ F(A^1,\ldots, A^k, \ldots , A^n)$$
His proof that the determinant is multilinear is in the linked image. It begins at: "1. Consider". (Note: he refers to the multilinearity condition as "property 1)".
Lang's proof of the multilinearity of the determinant
My problem is that I don't understand Lang's proof. Can someone please give me an alternative proof, or explain to me precisely how his proof shows that the determinant satisfies his definition of multilinearity? 

I imagine this is meant to be by induction. A linear combination of linear functions is again linear, and he's showing that every term of the sum is a linear function of the $k$-th column.
Inductively, when $j\ne k$, $\det(A_{ij})$ itself is a linear function of the $k$-th column (with its $j$-th entry removed, which is a linear operation) and $a_{ij}$ is a constant, while when $j=k$, the value $a_{ij}$ itself is evidently a linear function of the $k$-th column, while $\det(A_{ij})$ is a constant.
