Let in , where . Show that
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 ) 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)= ?