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!
 A: "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}
A: The general proof is not difficult.
From the definition of a determinant (sum of products), the expansion must be a polynomial in $x_1,x_2,\cdots x_n$, of degree $0+1+2+\cdots n-1=\dfrac{(n-1)n}2$, and the coefficient of every term is $\pm1$.
On another hand, the determinant cancels whenever $x_j=x_k$, so that the polynomial must be a multiple of
$$(x_1-x_2)(x_1-x_3)(x_1-x_4)\cdots(x_1-x_n)\\
(x_2-x_3)(x_2-x_4)\cdots(x_2-x_n)\\
(x_3-x_4)\cdots(x_3-x_n)\\
\cdots\\
(x_n-x_{n-1})$$ ($\dfrac{(n-1)n}2$ factors).
Hence the determinant has no other choice than being $\pm$ this product.

For the $3\times3$ case,
$$\begin{vmatrix} 1 & x_1 & x_1^2\\  1 &x_2  & x_2^2\\  1 & x_3 &
 x_3^2 \end{vmatrix}=
\begin{vmatrix} 1 & x_1 & x_1^2\\  0 &x_2-x_1  & x_2^2-x_1^2\\  0 & x_3-x_1 &
 x_3^2-x_1^2 \end{vmatrix}=\begin{vmatrix} x_2-x_1  & x_2^2-x_1^2\\ x_3-x_1 &
 x_3^2-x_1^2 \end{vmatrix}=(x_2-x_1)(x_3-x_1)\begin{vmatrix} 1&x_2+x_1 \\1& x_3+x_1 \end{vmatrix}=(x_2-x_1)(x_3-x_1)(x_3-x_2).$$
A: Using induction on $n$ is maybe the most convenient way. Let's proof that if we assume that the result is true for $n - 1$ then the result is true for $n$. So, consider the Vandermonde matrix
$$
V =\begin{pmatrix}
   1 & x_{1} & x_{1}^{2} \cdots        & x_{1}^{n-1} \\
   1 & x_{2} & x_{2}^{2} \cdots        & x_{2}^{n-1} \\
     &       &           \vdots           \\
   1 & x_{n -1} & x_{n - 1}^{2} \cdots & x_{n-1}^{n-1} \\
   1 & x_{n}    & x_{n}^{2}     \cdots & x_{n}^{n-1}
\end{pmatrix}
$$
Now consider $\det(V)$ as a polynomial of degree $n-1$ in the variable $x_{n}$. We can compute $\det(V)$ using the $n$-th column by means of the expansion of $\det(V)$ in terms of cofactors, so
$$
  \det(V) = a_{n-1} x_{n}^{n-1} + a_{n-2} x_{n}^{n-2} + \cdots + a_{0} \implies
  a_{n-1} =
  \det
  \begin{pmatrix}
   1 & x_{1} & x_{1}^{2} \cdots        & x_{1}^{n-2} \\
   1 & x_{2} & x_{2}^{2} \cdots        & x_{2}^{n-2} \\
     &       &           \vdots           \\
   1 & x_{n -1} & x_{n - 1}^{2} \cdots & x_{n-1}^{n-2}
\end{pmatrix}\\
\text{(Note that $a_{n-1}$ is the leading coeficient of the polynomial $\det(V)$)}
$$
In addition, $x_{1}, x_{2}, \dots, x_{n-1}$ are the roots of $\det(V)$, so we can write
$$
   \det(V) = a_{n-1}(x_{n}- x_{1}) (x_{n} - x_{2}) \cdots (x_{n} - x_{n-1}).
$$
By induction hypothesis we have the
$$
   a_{n-1} = \prod_{1 \leq i < j \leq n-1} (x_{i} - x_{j}).
$$
Hence,
$$
   \det(V) = \left[\prod_{1 \leq i < j \leq n-1}(x_{i} - x_{j})\right] (x_{n} - x_{1}) \dots (x_{n} - x_{n-1})
           = \prod_{1 \leq i < j \leq n} (x_{i} - x_{j}).
$$
