This is another way to look at it. $Ax = b$ where A is invertible. First we define the following matrix:

$$I_i(x) = [e_1\,\,e_2\,\, e_{i-1}\,\,x\,\,e_{i+1}\,\, ...\,\, e_n]$$
By the defenition of matrix multiplication:
$$AI_i(x) = [Ae_1\,\,Ae_2\,\, Ae_{i-1}\,\,Ax\,\,Ae_{i+1}\,\, ...\,\, Ae_n]$$
$$AI_i(x) = A_i(b)$$
$$det\,AI_i(x) = det\,A_i(b)$$

The determinant of the product of two matrices is the product of the determinants, so:

$$det\,A \,\cdot\, det\,I_i(x) = det\,A_i(b)$$

Let's now look at $det\,I_i(x)$, note that the determinants are the same because you can get rid of all the $x_j$ where $j\ne i$ by row reduction. Also $x$ is in the $i^{th}$ column:
$$det\,I_i(x) = det
\begin{bmatrix}
1 & x_1 & \cdots & 0 \\
0 & x_2 & 0 & 0 \\
\vdots & \vdots & \ddots & 0 \\
0 & x_n & 0 & 1 \\
\end{bmatrix}
= det
\begin{bmatrix}
1 & 0 & \cdots & 0 \\
0 & x_i & 0 & 0 \\
\vdots & \vdots & \ddots & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
= 1 \cdot 1 \cdot\cdot\cdot 1\cdot x_i \cdot 1\cdot\cdot\cdot1=x_i$$

$det\,I_i(x)$ is simply $x_i$, So we can say that:
$$x_i = \frac{det\,A_i(b)}{det\,A}$$