TITLE: Show that the determinant of a given matrix is equal to the product of the difference between all of its entries
QUESTION [0 upvotes]: Let $\alpha_1, \alpha_2, \dots, \alpha_n \in \mathbb{R}$, where $n\ge 2$. Show that
$$\left|\begin{matrix} 1& \alpha_1 & \alpha_1^2 & \dots & \alpha_1^{n-1} \\ 1 & \alpha_2 & \alpha_2^2 & \dots & \alpha_2^{n-1} \\ \dots & \dots & \dots & \dots & \dots \\ 1 & \alpha_n & \alpha_n^2 & \dots & \alpha_n^{n-1} \end{matrix}\right|= \prod_{1 \le i < j \le n}(\alpha_j - \alpha_i)$$
I tried to prove this by induction, so I first showed that the equation is true for $n=2$, I also checked that the equation is true for $n=3$, but then I am stuck on the inductive step.
So I assumed that the equation is true for $n=k$ and I wanted to show that then the equation will also be true for $n=k+1$, but I have no idea where to begin this prove.
Any help is appreciated, thanks in advance.

REPLY [1 votes]: First make elementary column operations so that the elements in the first row, columns $>1$, be $0$, be ginning with the rightmost column:

$C_n\leftarrow C_n -\alpha_1C_{n-1}$,
$C_{n-1}\leftarrow C_{n-1} -\alpha_1C_{n-2}$,

and so forth. Then use Laplace expansion along the first row: the determinant is the cofactor of the $1$ of the first row. You can factor each $\alpha_i-\alpha_1\enspace (i\ge 2)$ out of the rows of the cofactor and apply the inductive hypothesis to what remains.
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