Theorem: Value of Vandermonde Determinant/Formulation 1

Let $V_n$ be the '''Vandermonde determinant of order $n$''' defined as the following formulation:
{{:Definition:Vandermonde Determinant/Formulation 1}}
Its value is given by:
:$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$

Proof:
Proof by induction:
Let the Vandermonde determinant be presented in the following form:
:<nowiki>$V_n = \begin {vmatrix}
{x_1}^{n - 1} & {x_1}^{n - 2} & \cdots &    x_1 &      1 \\
{x_2}^{n - 1} & {x_2}^{n - 2} & \cdots &    x_2 &      1 \\
       \vdots &        \vdots & \ddots & \vdots & \vdots \\
{x_n}^{n - 1} & {x_n}^{n - 2} & \cdots &    x_n &      1 \\
\end {vmatrix}$</nowiki>
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_i - x_j}$
$\map P 1$ is true, as this just says $\begin {vmatrix} 1 \end {vmatrix} = 1$.
=== Basis for the Induction ===
$\map P 2$ holds, as it is the case:
:<nowiki>$V_2 = \begin {vmatrix}
a_1 & 1 \\
a_2 & 1
\end {vmatrix}$</nowiki>
which evaluates to $V_2 = a_1 - a_2$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds V_k = \prod_{1 \mathop \le i \mathop < j \mathop \le k} \paren {x_i - x_j}$
Then we need to show:
:$\ds V_{k + 1} = \prod_{1 \mathop \le i \mathop < j \mathop \le k + 1} \paren {x_i - x_j}$
=== Induction Step ===
This is our induction step:
Take the determinant:
:<nowiki>$V'_{k + 1} = \begin{vmatrix}
            x^k &            x^{k - 1} & \cdots &            x^2 &         x & 1 \\
        {x_2}^k &        {x_2}^{k - 1} & \cdots &        {x_2}^2 &       x_2 & 1 \\
         \vdots &               \vdots & \ddots &         \vdots &    \vdots & \vdots \\
 {x_{k + 1} }^k & {x_{k + 1} }^{k - 1} & \cdots & {x_{k + 1} }^2 & x_{k + 1} & 1
\end{vmatrix}$</nowiki>
where it is seen that $x_1$ has been replaced in $V_{k + 1}$ with $x$.
Let the Laplace Expansion Theorem for Determinants be used to expand $V'_{k + 1}$ in terms of the first row.
It can be seen that it is a polynomial in $x$ whose degree is no greater than $k$.
Let that polynomial be denoted $\map f x$.
Let any $x_r$ be substituted for $x$ in the determinant.
Then two of its rows will be the same.
From Square Matrix with Duplicate Rows has Zero Determinant, the value of such a determinant will be $0$.
Such a substitution in the determinant is equivalent to substituting $x_r$ for $x$ in $\map f x$.
Thus it follows that:
:$\map f {x_2} = \map f {x_3} = \cdots = \map f {x_{k + 1} } = 0$
So $\map f x$ is divisible by each of the factors $x - x_2, x - x_3, \ldots, x - x_{k + 1}$.
All these factors are distinct, otherwise the original determinant is zero.
So:
:$\map f c = \map C {x - x_2} \paren {x - x_3} \cdots \paren {x - x_k} \paren {x - x_{k + 1} }$
As the degree of $\map f x$ is no greater than $k$, it follows that $C$ is independent of $x$.
From the Laplace Expansion Theorem for Determinants, the coefficient of $x^k$ is:
:<nowiki>$\begin {vmatrix}
       {x_2}^{k - 1} & \cdots &        {x_2}^2 &       x_2 & 1 \\
              \vdots & \ddots &         \vdots &    \vdots & \vdots \\
{x_{k + 1} }^{k - 1} & \cdots & {x_{k + 1} }^2 & x_{k + 1} & 1
\end {vmatrix}$</nowiki>
By the induction hypothesis, this is equal to:
:$\ds \prod_{2 \mathop \le i \mathop < j \mathop \le k + 1} \paren {x_i - x_j}$
This must be our value of $C$.
So we have:
:$\ds \map f x = \paren {x - x_2} \paren {x - x_3} \dotsm \paren {x - x_k} \paren {x - x_{k + 1} } \prod_{2 \mathop \le i \mathop < j \mathop \le k + 1} \paren {x_i - x_j}$
Substituting $x_1$ for $x$, we retrieve the proposition $\map P {k + 1}$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_i - x_j}$
{{qed}}