From Serge Lang's Linear Algebra:
Let $x_1$, $x_2$, $x_3$ be numbers. Show that:
$$\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 & x_3^2 \end{vmatrix}=(x_2-x_1)(x_3-x_1)(x_3-x_2)$$
The matrix presented above seems to be the specific case of Vandermonde determinant:
$$ \begin{vmatrix} 1 & x_1 & ... & x_1^{n-1}\\ 1 &x_2 & ... & x_2^{n-1}\\ ... & ... & ... & ...\\ 1 & x_n & ... & x_n^{n-1} \end{vmatrix}=\prod_{i, j}(x_i - x_j), \forall (1 \leq i \leq n) \land (1 \leq j \leq n) $$
I'm trying to prove the specific case to then generalize it for arbitrary Vandermonde matrices.
My incomplete "proof"
Since determinant is a multilinear alternating function, it can be seen that adding a scalar multiple of one column (resp. row) to other column (resp. row) does not change the value (I omitted the proof to avoid too much text).
Thus considering that $x_1$ is a scalar, we can multiply each column but the last one of our specific Vandermonde matrix by $x_1$ and then starting from right to left subtract $n-1$th column from $n$:
$$\begin{vmatrix} 1 & x_1 & x_1^2\\ 1 &x_2 & x_2^2\\ 1 & x_3 & x_3^2 \end{vmatrix}=\begin{vmatrix} x_1 & 0 & 0 \\ x_1 & x_2 - x_1 & x^{2}_2 - x^{2}_1\\ x_1 & x_3 - x_1 & x^{2}_3 - x^{2}_1 \end{vmatrix}$$
Then using the expansion rule along the first row (since all the elements in it but $x_1$ are zero):
$$... =x_1\begin{vmatrix} x_2 - x_1 & x^{2}_2 - x^{2}_1\\ x_3 - x_1 & x^{2}_3 - x^{2}_1 \end{vmatrix}=(x_1x_2-x^2_1)(x^2_{3}-x^2_1)-(x^{2}_2x_1 - x^{3}_1)(x_3x_1 - x^2_1)$$
The first expansion seems interesting because it contains $x_2 - x_1$ and $x_3 - x_1$ (which are first two factors of specific Vandermonde matrix), but further expansion does not give satisfying results.
Question:
Is this a good simple start of inductively "proving" relation between Vandermonde matrix and its factors? If so what does it lack to show the complete result? Did I make mistake during evaluation?
Thank you!