Vandermonde Determinant
Let \(R\) be a commutative ring. Let \(a_1,a_2,... \in R\). Then
\[\begin{vmatrix}
1 & a_1 & ... & a_1^{n-1}
\\
\vdots & \vdots & \ddots & \vdots
\\
1 & a_n & ... & a_n^{n-1}
\end{vmatrix}
= { \prod_{1\le r< s\le n}(a_s - a_r) }\]
Proof: Assume the result holds for \(n\). Consider
\[\begin{vmatrix}
1 & a_1 & ... & a_1^n
\\
\vdots & \vdots & \ddots & \vdots
\\
1 & a_n & ... & a_{n+1}^n
\end{vmatrix}\]
This is equal to
\[\begin{vmatrix}
1 & a_1 & ... & a_1^{n-1} & f(a_1)
\\
\vdots & \vdots & \ddots & \vdots & \vdots
\\
1 & a_n & ... & a_{n+1}^{n-1} & f(a_{n+1})
\end{vmatrix}\]
for any monic \(f\in R[x]\) of degree \(n\), because this matrix can be obtained from the previous one via elementary column operations. In particular, if we set \(f = {\prod_{i=1}^{n}(x-a_i)}\) then \(f(a_1) = ... = f(a_n) = 0\) and \(f(a_{n+1}) = {\prod_{i=1}^n(a_{n+1}-a_i)}\). The result follows after using the inductive hypothesis.