I have the following problem:
Let $n\in\mathbb{N}$ and $A(n)$ an $(n\times n)$-matrix, $$ A(n)=\begin{pmatrix}a_{1,1} & a_{1,2} & \ldots & a_{1,n}\\ a_{2,1} & a_{2,2} & \ldots & a_{2,n}\\ \vdots & \ddots & \ldots & \vdots\\ a_{n,1} & a_{n,2} & \ldots & a_{n,n}\end{pmatrix} $$
To compute the determinant of $A(n)$, there is the famous Laplace expansion. I would like to expand repeatedly in the last row and always express everything in terms of the original entries of $A(n)$ (see example below):
In the first step, I want to expand in the $n$-th row.
In the next step, I want to expand in the $(n-1)$-th row and again express everything in terms of the original entries of $A(n)$
I want to repeat this until I end up with determinants of $(2\times 2$)-matrices.
Is it possible to get a formula for this? On the one hand, it seems to be complicated but on the other hand, it seems to be "only" a problem of bookkeeping and indices.
Example: $n=4$:
Step 1: Expand in the 4-th row $$ \lvert A(4)\rvert = \sum_{i=1}^4 a_{4,i}\cdot (-1)^{4+i} \lvert A_{4,i}\rvert\tag{*}, $$ where $A_{4,i}$ is the minor of $A(4)$ which one gets when deleting the 4-th row and the i-th column of $A(n)$.
Step 2: Compute $\lvert A_{4,i}\rvert, i=1,2,3,4$ again by expanding in the last row and express everything in terms of the original matrix $A(4)$: $$ \lvert A_{4,i}\rvert = \sum_{j<i}a_{3,j}\cdot (-1)^{3+j}\lvert (A_{4,i})_{3,j}\rvert + \sum_{j>i}^n a_{3,j}\cdot (-1)^{3+(j-1)}\lvert (A_{4,i})_{3,j-1}\rvert $$
Step 3: Repeat this until we end up with only $(2\times 2)$-matrices.
In the end: Plug everything into $(*)$.
If you repeat this process, you should get a formula of the form: $det(A(n)) = \sum_\sigma sign (\sigma)a_{1,\sigma(1)}...a_{n,\sigma(n)} $ where $\sigma$ ranges over all permutations of $n$ elements. Some textbooks give this as the definition of the determinant.
This formula follows only from the fact that the determinant is linear in all of the columns of the matrix, changes sign when you swap two columns, and has value 1 on the identity. The first two of these properties are summed up by saying that the determinant is multilinear and alternating.