Value of Vandermonde Determinant/Formulation 1/Proof 2

Proof
Proof by induction:

Let the Vandermonde Determinant be presented in the form as defined by Mirsky:

Let $V_n = \begin{vmatrix} a_1^{n-1} & a_1^{n-2} & \cdots & a_1 & 1 \\ a_2^{n-1} & a_2^{n-2} & \cdots & a_2 & 1 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ a_n^{n-1} & a_n^{n-2} & \cdots & a_n & 1 \\ \end{vmatrix}$

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i - a_j}\right)$

$P(1)$ is true, as this just says $\begin{vmatrix} 1 \end{vmatrix} = 1$.

Basis for the Induction
$P(2)$ holds, as it is the case:
 * $V_2 = \begin{vmatrix}

a_1 & 1 \\ a_2 & 1 \end{vmatrix}$ 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 $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle V_k = \prod_{1 \mathop \le i \mathop < j \mathop \le k} \left({a_i - a_j}\right)$

Then we need to show:
 * $\displaystyle V_{k+1} = \prod_{1 \mathop \le i \mathop < j \mathop \le k+1} \left({a_i - a_j}\right)$

Induction Step
This is our induction step:

Take the determinant:
 * $V_{k+1} = \begin{vmatrix}

x^k & x^{k-1} & \cdots & x^2 & x & 1 \\ a_2^k & a_2^{k-1} & \cdots & a_2^2 & a_2 & 1 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ a_{k+1}^k & a_{k+1}^{k-1} & \cdots & a_{k+1}^2 & a_{k+1} & 1 \end{vmatrix}$

Let the Expansion Theorem for Determinants‎ be used to expand $V_n$ 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 $f \left({x}\right)$.

Let any $a_r$ 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 $a_r$ for $x$ in $f \left({x}\right)$.

Thus it follows that:
 * $f \left({a_2}\right) = f \left({a_3}\right) = \ldots = f \left({a_{k+1}}\right) = 0$

So $f \left({x}\right)$ is divisible by each of the factors $x - a_2, x - a_3, \ldots, x - a_{k+1}$.

All these factors are distinct, otherwise the original determinant is zero.

So:
 * $f \left({x}\right) = C \left({x - a_2}\right) \left({x - a_3}\right) \cdots \left({x - a_k}\right) \left({x - a_{k+1}}\right)$

As the degree of $f \left({x}\right)$ is no greater than $k$, it follows that $C$ is independent of $x$.

From the Expansion Theorem for Determinants‎, the coefficient of $x^k$ is:
 * $\begin{vmatrix}

a_2^{k-1} & \cdots & a_2^2 & a_2 & 1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ a_{k+1}^{k-1} & \cdots & a_{k+1}^2 & a_{k+1} & 1 \end{vmatrix}$.

By the induction hypothesis, this is equal to:
 * $\displaystyle \prod_{2 \mathop \le i \mathop < j \mathop \le k+1} \left({a_i - a_j}\right)$

This must be our value of $C$.

So we have:
 * $\displaystyle f \left({x}\right) = \left({x - a_2}\right) \left({x - a_3}\right) \cdots \left({x - a_k}\right) \left({x - a_{k+1}}\right) \prod_{2 \mathop \le i \mathop < j \mathop \le k+1} \left({a_i - a_j}\right)$

Substituting $a_1$ for $x$, we retrieve the proposition $P \left({k+1}\right)$.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i - a_j}\right)$