I can follow the proof given in (2nd proof, or the induction proof), until the sentence: "From the Expansion Theorem for Determinants, we can see that the coefficient of $x_k$ is:".
I don't understand why it is only needed the determinant of the cofactor of $x_k$, since the calculation of a determinant involves taking one row and the multiplication of the cofactors in the chosen row.
On
It's talking about how you work out determinants you can "expand" them into a sum of 2x2 determinants
It's what I did here ( Vandermonde determinant by induction ) 2 years ago.
It has been shown that the determinant is something like:
$$V_{k+1}(x) = f(x) = C(x-a_2)(x-a_3)\ldots(x-a_k)(x-a_{k+1})$$
Note that if you would expand this it would be of the form:
$$f(x) = Cx^k + B_{k-1}x^{k-1}+\ldots$$
Now consider the expansion theorem for the first row:
$$V_{k+1}(x) = \begin{vmatrix} a_2^{k-1} & \cdots & a_2^2 & a_2 & 1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ a_{k+1}^{k-1} & \cdots & a_{k+1}^2 & a_{k+1} & 1 \end{vmatrix}x^k +B_{k-1}x^{k-1}+\ldots$$
Where $B_{i}$ is some constant.