The general Vandermonde determinant

Base case $n=2$. $V(a_1,a_2)=a_2-a_1$.

Inductive step. Subtract column 1 from columns $2,\ldots,n$:
$$V = \begin{vmatrix} 1 & 0 & \cdots & 0 \\ a_1 & a_2-a_1 & \cdots & a_n-a_1 \\ a_1^2 & a_2^2-a_1^2 & \cdots & a_n^2-a_1^2 \\ \vdots & \vdots & & \vdots \\ a_1^{n-1} & a_2^{n-1}-a_1^{n-1} & \cdots & a_n^{n-1}-a_1^{n-1} \end{vmatrix}.$$
Expand along the first row. For each column $j\ge2$, factor $a_j-a_1$:
$$V = \left(\prod_{j=2}^n(a_j-a_1)\right)\cdot\begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_1+a_2 & a_1+a_3 & \cdots & a_1+a_n \\ \vdots & \vdots & & \vdots \\ \sum_{k=0}^{n-2}a_1^{n-2-k}a_j^k & \cdots & & \end{vmatrix}.$$
By column operations (subtracting $a_1$ times the previous column), the remaining determinant reduces to $V(a_2,\ldots,a_n)$.

Induction. By the induction hypothesis, $V(a_2,\ldots,a_n)=\prod_{2\le i<j\le n}(a_j-a_i)$.

Conclusion.
$$V(a_1,\ldots,a_n) = \prod_{j=2}^n(a_j-a_1)\prod_{2\le i<j\le n}(a_j-a_i) = \prod_{1\le i<j\le n}(a_j-a_i).$$