Theorem: Laplace's Expansion Theorem

Let $D$ be the determinant of order $n$.
Let $r_1, r_2, \ldots, r_k$ be integers such that:
:$1 \le k < n$
:$1 \le r_1 < r_2 < \cdots < r_k \le n$
Let $\map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$ be an order-$k$ minor of $D$.
Let $\map {\tilde D} {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$ be the cofactor of $\map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$.
Then:
:$\ds D = \sum_{1 \mathop \le u_1 \mathop < \cdots \mathop < u_k \mathop \le n} \map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k} \, \map {\tilde D} {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$
A similar result applies for columns.

Proof:
Let us define $r_{k + 1}, r_{k + 2}, \ldots, r_n$ such that:
:$1 \le r_{k + 1} < r_{k + 2} < \cdots < r_n \le n$
:$\rho = \tuple {r_1, r_2, \ldots, r_n}$ is a permutation on $\N^*_n$.
Let $\sigma = \tuple {s_1, s_2, \ldots, s_n}$ be a permutation on $\N^*_n$.
Then by Permutation of Determinant Indices we have:
{{begin-eqn}}
{{eqn | l = D
      | r = \sum_\sigma \map \sgn \rho \, \map \sgn \sigma \prod_{j \mathop = 1}^n a_{\map \rho j \, \map \sigma j}
      | c = 
}}
{{eqn | r = \sum_\sigma \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + s_i} } \map \sgn {\map \rho {r_1, \ldots, r_k} } \, \map \sgn {\map \sigma {s_1, \ldots, s_k} } \map \sgn {\map \rho {r_{k + 1}, \ldots, r_n} } \, \map \sgn {\map \sigma {s_{k + 1}, \ldots, s_n} } \prod_{j \mathop = 1}^n a_{\map \rho j \, \map \sigma j}
      | c = 
}}
{{end-eqn}}
We can obtain all the permutations $\sigma$ exactly once by separating the numbers $1, \ldots, n$ in all possible ways into a set of $k$ and $n - k$ numbers.
We let $\tuple {s_1, \ldots, s_k}$ vary over the first set and $\tuple {s_{k + 1}, \ldots, s_n}$ over the second set.
So the summation over all $\sigma$ can be replaced by:
:$\tuple {u_1, \ldots, u_n} = \map \sigma {1, \ldots, n}$
:$u_1 < u_2 < \cdots < u_k, u_{k + 1} < u_{k + 2} < \cdots < u_n$
:$\tuple {s_1, \ldots, s_k} = \map \sigma {u_1, \ldots, u_k}$
:$\tuple {s_{k + 1}, \ldots, s_n} = \map \sigma {u_{k + 1}, \ldots, u_n}$
Thus we get:
{{begin-eqn}}
{{eqn | l = D
      | r = \sum_{\map \sigma {u_1, \ldots, u_n} } \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + u_i} } \sum_{\map \sigma {u_1, \ldots, u_k} } \, \map \sgn {\map \rho {r_1, \ldots, r_k} } \, \map \sgn {\map \sigma {s_1, \ldots, s_k} } \prod_{j \mathop = 1}^k a_{\map \rho j \, \map \sigma j}
      | c = 
}}
{{eqn | o = \times
      | r = \sum_{\map \sigma {u_{k + 1}, \ldots, u_n} } \map \sgn {\map \rho {r_{k + 1}, \ldots, r_n} } \, \map \sgn {\map \sigma {s_{k + 1}, \ldots, s_n} } \prod_{j \mathop = k + 1}^n a_{\map \rho j \, \map \sigma j}
      | c = 
}}
{{eqn | r = \sum_{\map \sigma {u_1, \ldots, u_n} } \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + u_i} } \begin {vmatrix} a_{r_1 u_1} & \cdots & a_{r_1 u_k} \\ \vdots & \ddots & \vdots \\ a_{r_k u_1} & \cdots & a_{r_k u_k} \end {vmatrix} \times \begin {vmatrix} a_{r_{k + 1} u_{k + 1} } & \cdots & a_{r_{k + 1} u_n} \\ \vdots & \ddots & \vdots \\ a_{r_n u_{k + 1} } & \cdots & a_{r_n u_n} \end {vmatrix}
      | c = 
}}
{{eqn | r = \sum_{\map \sigma {u_1, \ldots, u_n} } \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + u_i} } \map D {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \times \map D {r_{k + 1}, \ldots, r_n \mid u_{k + 1}, \ldots, u_n}
      | c = 
}}
{{eqn | r = \sum_{\map \sigma {u_1, \ldots, u_n} } \map D {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \times \map {\tilde D} {r_1, \ldots, r_k \mid u_1, \ldots, u_k}
      | c = 
}}
{{eqn | r = \sum_{1 \mathop \le u_1 \mathop < \cdots \mathop < u_k \mathop \le n} \map D {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \, \map {\tilde D} {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \sum_{u_{k + 1}, \ldots, u_n} 1
      | c = 
}}
{{end-eqn}}
That last inner sum extends over all integers which satisfy:
:$\tuple {u_1, \ldots, u_n} = \map \sigma {1, \ldots, n}$
:$u_1 < u_2 < \cdots < u_k, u_{k + 1} < u_{k + 2} < \cdots < u_n$
But for each set of $u_1, \ldots, u_k$, then the integers $u_{k + 1}, \ldots, u_n$ are clearly uniquely determined.
So that last inner sum equals 1 and the theorem is proved.
{{Explain|I'm not too happy about this, it seems a bit handwavey and imprecise. I'm going to have to revisit it.}}
The result for columns follows from Determinant of Transpose. 
{{qed}}
{{Proofread}}