How to apply Vandemonde determinant to show nilpotent here?
Solution 1:
From part (i), we know that $$ \lambda_1^m + \cdots + \lambda_n^m = 0 $$ For all $m$ from $1$ to $n$. Now, I'm not quite sure how they were planning on using the Vandermonde determinant, but a quick application of Newton's identities allows you to deduce that all $\lambda_i = 0$.