Inductively simplify specific Vandermonde determinant 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!

"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)
" is right. But
$$
\begin{vmatrix} 1 & x_1 & x_1^2\\  1 &x_2  & x_2^2\\  1 & x_3 &
 x_3^2 \end{vmatrix}
\neq
\begin{vmatrix}
x_1 & 0 & 0 \\ 
x_1 & x_2 - x_1 & x^{2}_2 - x^{2}_1\\ 
x_1 & x_3 - x_1 & x^{2}_3 - x^{2}_1
\end{vmatrix}
\neq
(x_1x_2-x^2_1)(x^2_{3}-x^2_1)-(x^{2}_2x_1 - x^{3}_1)(x_3x_1 - x^2_1)
$$
Remember that when you multiply a row or a column by $\lambda$, the determinant is multiplied by $\lambda$. And be careful when distributing $x_1$. We have
\begin{align}
\begin{vmatrix} 1 & x_1 & x_1^2\\  1 &x_2  & x_2^2\\  1 & x_3 &
 x_3^2 \end{vmatrix}
&=
x_1
\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}\\
&= x_1^2
\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_1^2((x_2 - x_1)(x^{2}_3 - x^{2}_1) - (x^{2}_2 - x^{2}_1)(x_3 - x_1))\\
&\neq (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)
\end{align}
Keep in mind that we are trying to have the simplest possible factors.
Here, you can do
\begin{align}
\begin{vmatrix} 1 & x_1 & x_1^2\\  1 &x_2  & x_2^2\\  1 & x_3 &
 x_3^2 \end{vmatrix}&=_{L_3 \leftarrow L_3 - L_2 \text{ and } L_2 \leftarrow L_2 - L_1}
\begin{vmatrix} 1 & x_1 & x_1^2\\  0 &x_2  -x_1& (x_2 - x_1)(x_2+x_1)\\  0 & x_3 - x_2 &
 (x_3 - x_2)(x_3+x_2) \end{vmatrix}\\
&=_{L_3 \leftarrow L_3 - L_2} (x_2 - x_1)(x_3-x_2)
\begin{vmatrix} 1 & x_1 & x_1^2\\  0 &1& x_2 + x_1\\  0 & 0 &
 x_3 -x_1 \end{vmatrix}\\
&=(x_2 - x_1)(x_3-x_2)(x_3-x_1)
\end{align}
