How to apply Vandemonde determinant to show nilpotent here? This is question 5.5 from Erdmann & Wildon (paraphrased):
Consider 2 square matrices over complex number $A,B$. Let $C=AB-BA$. Assume that $AC=CA$ and $BC=CB$.
(i) Show that $trace(C^{m})=0$ for all $m\geq 1$.
(ii) Let $n\times n$ be the dimension of those matrices. Let $\lambda_{1},\ldots,\lambda_{n}$ be the eigenvalues of $C$. Prove that $\lambda_{i}=0$ for all $i$, and deduce that $C^{k}=0$ for some $k$ (hint: use Vandermonde determinant).
(note: this was written as a Lie algebra question, but since the question explicitly ask to avoid the use of invariance lemma, it looks clearly more like a linear algebra question, which is why I rephrased it into a linear algebra question)
I already solved (i). But I have no clues how to do (ii). The question hinted at using Vandermonde determinant. This is the first time I heard of it, and after looking it up I still gain no intuitions about it to help me handle this question.
Thus I am looking for help in solving (ii). Thank you for your help.
 A: 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 $\lambda_i = 0$.
A: Continuing from Ben Grossmann's answer, using the Vandermonde determinant:
In general, I claim that for $c_1,\dots,c_n\in\mathbb R_{>0}$ and $\lambda_1,\dots,\lambda_n\in\mathbb C$, if for any $m>0$ you have
$$c_1\lambda_1^m+\dots+c_n\lambda_n^m=0,$$
then $\lambda_1=\dots=\lambda_n=0$.
Proof. It is clear for $n=1$. Assume $n>1$, and that the theorem is true for $n-1$.
Now the vectors $(\lambda_1\dots\lambda_1^n),\dots,(\lambda_n\dots\lambda_n^n)$ are linearly dependent, so the determinant
$$\det\begin{pmatrix}\lambda_1&\dots&\lambda_n\\&\ddots&\\\lambda_1^n&\dots&\lambda_n^n\end{pmatrix}=\lambda_1\cdots\lambda_n\prod_{i<j}(\lambda_i-\lambda_j)$$
must be $0$. If some $\lambda_i=0$ then apply the inductive hypothesis on $c_1,\dots,c_{i-1},c_{i+1},\dots,c_n$ and $\lambda_1,\dots,\lambda_{i-1},\lambda_{i+1},\dots,\lambda_n$.
If $\lambda_i=\lambda_j$ for $i<j$, say when $i=1$ and $j=2$ (without loss), then apply the inductive hypothesis on $c_1+c_2,c_3\dots,c_n$ and $\lambda_2,\lambda_3,\dots,\lambda_n$.
