The formula $A \cdot (B \times C) = \textrm{Det}(A,B,C)$ shows this the cross product can be thought of as the transpose of the linear map $\textrm{Det}(\cdot,B,C)$.
Using the notation of riemannian geometry (hodge star, sharps, and flats) another way to say this is that $A \times B=\star(A^\flat \wedge B^\flat)^\sharp$. This is the connection between the cross product and exterior product you were looking for.
Hmm. I actually just remembered that I answered a similar question here. I am thinking of the entries as columns rather than rows in the determinant here (because that is how my brain works), but I think you will see the connection.
You may also be interested in this question on MO.
If you learn index (summation) notation, then you'll be able to prove it without writing out every term:
Consider an arbitrary $p$th coordinate of the vector $A\times (B\times C)$. Then
$$\begin{align} [A\times (B\times C)]_p &= \varepsilon_{pqr}A_q(B\times C)_r \\ &= \varepsilon_{pqr}A_q\varepsilon_{rst}B_sC_t \\ &= \varepsilon_{pqr}\varepsilon_{rst}A_qB_sC_t \\ &= \varepsilon_{rpq}\varepsilon_{rst}A_qB_sC_t \\ &= (\delta_{ps}\delta_{qt}-\delta_{pt}\delta_{qs})A_qB_sC_t \\ &= \delta_{ps}\delta_{qt}A_qB_sC_t -\delta_{pt}\delta_{qs}A_qB_sC_t \\ &= A_tB_pC_t - A_sB_sC_p \\ &= (A\cdot C)B_p - (A\cdot B)C_p \\ &= [(A\cdot C)B - (A\cdot B)C]_p\end{align}$$
Because this holds for an arbitrary $p$th coordinate of both vectors, it holds for the vectors themselves.$\ \ \ \ \ \square$
You still seem to be stuck in your derivation, so here's how to do it:
$$(a_x,a_y,a_z)\times \left[(b_x,b_y,b_z)\times(c_x,c_y,c_z)\right] = (a_x,a_y,a_z)\times (b_yc_z-b_zc_y,b_zc_x-b_xc_z,b_xc_y-b_yc_x) \\ = \left(\color{red}{a_y(b_xc_y-b_yc_x) - a_z(b_zc_x-b_xc_z)}, \color{purple}{a_z(b_yc_z-b_zc_y)-a_x(b_xc_y-b_yc_x)}, \color{blue}{a_x(b_zc_x-b_xc_z)-a_y(b_yc_z-b_zc_y)}\right)$$
If you compare this to what you got on the RHS you see that it is the same.
Your problem is just that you're writing down the negative of the middle component of each cross product. If you had just switched the signs of all the affected terms, you'd have gotten the same thing.
Best Answer
The determinant of a $3\times3$ matrix can be viewed as the triple product of its columns (or rows): $$ \begin{align} \det\begin{bmatrix} x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3 \end{bmatrix} &= \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} \times \begin{bmatrix} y_1\\ y_2\\ y_3 \end{bmatrix} \cdot \begin{bmatrix} z_1\\ z_2\\ z_3 \end{bmatrix}\\ &= \begin{bmatrix} (x\times y)_1\\ (x\times y)_2\\ (x\times y)_3 \end{bmatrix} \cdot \begin{bmatrix} z_1\\ z_2\\ z_3 \end{bmatrix}\tag{1} \end{align} $$ If we replace $\begin{bmatrix} z_1\\ z_2\\ z_3 \end{bmatrix}$ in $(1)$ by $\begin{bmatrix} \boldsymbol{i}\\ \boldsymbol{j}\\ \boldsymbol{k} \end{bmatrix}$, we get $$ \begin{align} \det\begin{bmatrix} x_1&y_1&\boldsymbol{i}\\ x_2&y_2&\boldsymbol{j}\\ x_3&y_3&\boldsymbol{k} \end{bmatrix} &= \begin{bmatrix} (x\times y)_1\\ (x\times y)_2\\ (x\times y)_3 \end{bmatrix} \cdot \begin{bmatrix} \boldsymbol{i}\\ \boldsymbol{j}\\ \boldsymbol{k} \end{bmatrix}\\[6pt] &=(x\times y)_1\boldsymbol{i}+(x\times y)_2\boldsymbol{j}+(x\times y)_3\boldsymbol{k}\\[18pt] &=x\times y\tag{2} \end{align} $$