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