Argument that Vandermonde matrix's determinant has $n-1$ distinct roots

Question

det(Vandermonde) = $\left|\begin{array}{ccccc}1 & x & x^2 & ... & x^{n-1} \\1 & a_2 & a^{2}_{2} & ... & a^{n-1}_{2} \\1 & ... & ... & ... & ... \\1 & a_n & a^2_{n} & ... & a^{n-1}_n\end{array}\right|$.

Let $P(x)$ be the polynomial in $x$ that you get from expanding along first row minors.

How do you show that $a_2, \dots, a_n$ are $n-1$ distinct roots of $P(x)$? How do I know that $a_2, \dots, a_n$ are distinct roots of this polynomial?

Answer 1

Hint: If two rows (or columns) of a matrix are the same then its determinant is 0.