Edit: I'm genuinely not sure why this has gotten little activity. If somebody knows, please tell me, so I can rework it.
As a note: I am a purist, and really want to see a proof of this, but I've also got course obligations to fulfill, and so not much time. I understand that the answer involves group theory, where my backing is weak, so more complete answers would really help me.
Let $A$ be an $n\times n$ matrix of real numbers. Define the $(i, j)$th minor $M_{(i, j)}$ of $A$ to be the determinant of the $(n-1) \times (n-1)$ matrix obtained by removal of the $i$th row and $j$th column from $A$ - e.g.,
$A = \begin{bmatrix}1 & 4 & 2 \\9 & 4 & 6 \\8 & 3 & 1 \end{bmatrix} \implies M_{(1, 1)} = \begin{vmatrix}4 & 6 \\3 & 1 \end{vmatrix}, \text{ }M_{(2,1)} = \begin{vmatrix} 4 & 2 \\ 3 & 1 \end{vmatrix}, \mathrm{etc.}$
Define the $(i,j)$th cofactor $C_{(i, j)}$ by $C_{(i, j)} = A_{(i,j)}(-1)^{i + j}M_{(i,j)}$ - i.e., multiply $A_{(i,j)}M_{(i,j)}$ by a sign corresponding to the $(i,j)$th entry of the matrix
$\mathrm{sgn} = \begin{bmatrix} + & - & + & \cdots \\- & + & - & \cdots \\ + & - & + &\cdots \\ \vdots & \vdots & \vdots & \ddots \end{bmatrix};$
thus,
$A = \begin{bmatrix}1 & 4 & 2 \\9 & 4 & 6 \\8 & 3 & 1 \end{bmatrix} \implies C_{(1, 1)} = 1 \times \begin{vmatrix}4 & 6 \\3 & 1 \end{vmatrix}, \text{ }C_{(2,1)} = -9 \times \begin{vmatrix} 4 & 2 \\ 3 & 1 \end{vmatrix}.$
Finally, define the cofactor expansion $D_i(A)$ across the (i)th row of $A$ by $D_i(A) = \sum_{k = 1}^n C_{(i, k)}$. For example, we get
\begin{eqnarray} D_1 \Bigg( \begin{bmatrix}1 & 4 & 2 \\9 & 4 & 6 \\8 & 3 & 1 \end{bmatrix} \Bigg) = C_{(1, 1)} + C_{(1, 2)} + C_{(1, 3)} = \begin{vmatrix}4 & 6 \\3 & 1 \end{vmatrix} - 4 \begin{vmatrix} 9 & 6 \\ 8 & 1 \end{vmatrix} + 2\begin{vmatrix}9 & 4 \\8 & 3 \end{vmatrix} = 132 \\ D_2 \Bigg( \begin{bmatrix}1 & 4 & 2 \\9 & 4 & 6 \\8 & 3 & 1 \end{bmatrix} \Bigg) = C_{(2, 1)} + C_{(2, 2)} + C_{(2, 3)} = -9 \begin{vmatrix}4 & 2 \\3 & 1 \end{vmatrix} + 4 \begin{vmatrix} 1 & 2 \\ 8 & 1 \end{vmatrix} - 6\begin{vmatrix}1 & 4 \\8 & 3 \end{vmatrix} = 132 \\ D_3 \Bigg( \begin{bmatrix}1 & 4 & 2 \\9 & 4 & 6 \\8 & 3 & 1 \end{bmatrix} \Bigg) = C_{(3, 1)} + C_{(3, 2)} + C_{(3, 3)} = 8 \begin{vmatrix}4 & 2 \\4 & 6 \end{vmatrix} - 3 \begin{vmatrix} 1 & 2 \\ 9 & 6 \end{vmatrix} +\begin{vmatrix}1 & 4 \\9 & 4 \end{vmatrix} = 132. \end{eqnarray}
The equality of each of these expansions serves as an example of the theorem I want to see proven:
Theorem: For any row $\rho_i$ (or column, with a similar definition to that for rows), the cofactor expansion $D_i(A)$ of a given matrix $A$ has the same numerical value.
Some notes:
(1) (Yes, I do realize that all the $\TeX$ belies my claim to not have much time. What I mean is that I could use some instruction - as a way of saving quite a bit of valuable time - since I've never taken a course in algebra before.)
(2) I admit that I don't really understand the Leibniz formula, so I don't actually have an immediate definition of determinant to fall back on inductively. One way around is to prove the cofactor theorem inductively on the size of $n$. Since every property of the determinant follows (easily) from the cofactor theorem, the above theorem is all I need to have proof of at the moment.