Let $\displaystyle a_{1},a_{2},...,a_{n} $ in $\displaystyle R $ , where $\displaystyle n \geq 2$ . Show that

$\displaystyle \left| \begin{array} {ccccc}1 & a_{1} & a_{1}^{2} & ... & a_{1}^{n-1} \\ 1 & a_{2} & a_{2}^{2} & ... & a_{2}^{n-1} \\ ... & ... & ... & ... & ... \\ 1 & a_{n} & a_{n}^{2} & ... & a_{n}^{n-1} \end{array} \right| $ = $\displaystyle \prod_{1 \leq i < j \leq n} (a_{j}-a_{i})$.

Suggestion: Do it by induction.

Now clearly this is true for n=2. Also, it can quite easily be shown for n=3.

Now following the protocol for a proof by induction, we assume that it is true for n=k. Now it must somehow be possible to show that it is true for n=k+1 which would complete the proof.

For now all I have is that the matrix for n=k (for which we know the determinant is $\displaystyle \prod_{1 \leq i < j \leq k} (a_{j}-a_{i})$) is the matrix for n=k+1 minus the k+1th row and the k+1th column. Is there any way this can be used to prove the expected result that det(matrix for n=k+1)= $\displaystyle \prod_{1 \leq i < j \leq k+1} (a_{j}-a_{i})$ ?

Thanks!!