Proving a formula for the determiante in a special matrix

[rosh300]

Homework Statement

Let a₁, a₂ are real numbers, where n > 1 show that: determinant of:

| 1 a₁ a²₁ ... ... a^n-1₁ |
| 1 a₂ a²₂ ... ... a^n-1₂ |
:
:
| 1 a_n a²_n ... ... a^n-1_n |


= \prod (a_j - a_i)
_1\leqi<j<n

Homework Equations

if you row reduce a matrix the determinate is the product of the leading diagonal(previous question was finding determinate of matrices by row reducing them)

The Attempt at a Solution

tried using induction but get stuck very quickly.
i got RHS =
= \Pi_0<i<j<n (a_j - a_i) \Pi_0<k<n (a_n - a_k)

 

[tiny-tim]

show that: determinant of …
= \prod (a_j - a_i)
_1\leqi<j<n
…
i got RHS =
= \Pi_0<i<j<n (a_j - a_i) \Pi_0<k<n (a_n - a_k)

Hi rosh300!

But that's the answer, isn't it?

The first part is all non-identical pairs up to n-1, and the second part is all non-identical pairs of which the higher is n.