Actually it is about the question of n-linear function, but it is so relevant to the determinant formula.
Here is the notation of the theorem.
If $n>1$ and $A$ is an $n \times n$ matrix over $K$, we let $A(i|j)$ denote the $(n-1) \times (n-1)$ matrix obtained by deleting the $i$th row and $j$the column of $A$. If $D$ is an $(n-1)$-linear function and $A$ is an $n \times n$ matrix, we put $D_{ij}(A)= D[A(i|j)]$.
I want to prove the following theorem.
Let $n>1$ and let $D$ be an alternating $(n-1)$-linear function on $(n-1)\times (n-1)$ matrices over $K$. For each $j$, $1 \leq j \leq n$, the function $E_j$ defined by $$E_j(A) = \sum_{i=1}^n (-1)^{i+j} A_{ij} D_{ij}(A)$$ is an alternating $n$-linear function on $n \times n$ matrices $A$. If $D$ is a determinant function, so is each $E_j$.
The proof in my text says that "$D_{ij}(A)$ is independent of the $i$th row of $A$. Since $D$ is $(n-1)$-linear, it is clear that $D_{ij}$ is linear as a function of any row except row $i$. Therefore $A_{ij}D_{ij}(A)$ is an $n$-linear function of $A$."
I don't understand that $A_{ij}D_{ij}$ is $n$-linear function. Because $D_{ij}$ is not defined by the row $i$, and the column $j$, it is clearly $(n-1)$-linear function not $n$-linear function.
Help me to understand this.