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  "body": "\n## Proof\n\nWe argue by induction on $n$.\n\n1. For $x_1, \\dots, x_n \\in K$, write $V(x_1, \\dots, x_n)$ for the matrix with entries $x_i^{\\,j-1}$, and $P(x_1, \\dots, x_n) = \\prod_{1 \\le i \u003c j \\le n} (x_j - x_i)$.\n2. Base case $n = 1$: $V(x_1) = (1)$. By the [[Leibniz formula|016PE4VZ8Q]] with one term, its determinant is $1$. The product $P(x_1)$ is empty, so it is also $1$.\n3. Induction step: let $n \\ge 2$, and assume $\\det V(y_1, \\dots, y_{n-1}) = P(y_1, \\dots, y_{n-1})$ for all $y_1, \\dots, y_{n-1} \\in K$.\n4. Column operations: let $M'$ result from a square matrix $M$ when $c$ times column $k$ is added to column $l$, with $k \\ne l$.\n5. Then the [[transpose|0W809K6AW0]] $M'^T$ results from $M^T$ when $c$ times row $k$ is added to row $l$.\n6. By the rule for [[adding a multiple of a row|01ZTQBK64Q]] and the rule for the [[determinant of a transpose|0ZZQ2E1MTE]], $\\det M' = \\det M'^T = \\det M^T = \\det M$.\n7. Let $V = V(x_1, \\dots, x_n)$. For $j = n, n-1, \\dots, 2$, in this order, subtract $x_1$ times column $j - 1$ from column $j$. Call the result $V'$.\n8. When column $j$ is changed, column $j - 1$ has not been changed yet. So column $j$ of $V'$ has entries $x_i^{\\,j-1} - x_1 x_i^{\\,j-2} = (x_i - x_1)\\, x_i^{\\,j-2}$ for $j \\ge 2$.\n9. Column $1$ of $V'$ is unchanged, with all entries $1$. By step 6, applied $n - 1$ times, $\\det V' = \\det V$.\n10. Row $1$ of $V'$ is $(1, 0, \\dots, 0)$, since $x_1 - x_1 = 0$.\n11. By the [[Laplace expansion along row $1$|0C08BQ3M05]], $\\det V' = 1 \\cdot (-1)^{1+1} M'_{11} = M'_{11}$, where $M'_{11}$ is a first [[minor|0SD528P0EK]] of $V'$.\n12. $M'_{11} = \\det B$, where $B$ is the $(n-1) \\times (n-1)$ matrix with entries $B_{rs} = (x_{r+1} - x_1)\\, x_{r+1}^{\\,s-1}$ for $1 \\le r, s \\le n-1$.\n13. Let $W = V(x_2, \\dots, x_n)$, with entries $W_{rs} = x_{r+1}^{\\,s-1}$. Row $r$ of $B$ is $(x_{r+1} - x_1)$ times row $r$ of $W$.\n14. By the rule for [[scaling a row|009CY8JXFX]], applied to each of the $n - 1$ rows, $\\det B = \\prod_{r=1}^{n-1} (x_{r+1} - x_1) \\cdot \\det W$.\n15. By the induction hypothesis, $\\det W = P(x_2, \\dots, x_n) = \\prod_{2 \\le i \u003c j \\le n} (x_j - x_i)$.\n16. By steps 9 to 15, $\\det V = \\prod_{j=2}^{n} (x_j - x_1) \\cdot \\prod_{2 \\le i \u003c j \\le n} (x_j - x_i)$.\n17. The pairs $i \u003c j$ with $i = 1$ give the first product, and the pairs with $i \\ge 2$ give the second. So $\\det V = \\prod_{1 \\le i \u003c j \\le n} (x_j - x_i)$.\n\n## Idea\n\nSubtracting $x_1$ times each column from the next clears the first row except for a $1$. What is left has a factor $x_i - x_1$ in each row and is otherwise a smaller Vandermonde matrix. Induction then collects one factor $x_j - x_i$ for each pair $i \u003c j$.\n",
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