TITLE: $\operatorname{tr}(A)=\operatorname{tr}(A^{2})= \ldots = \operatorname{tr}(A^{n})=0$ implies $A$ is nilpotent
QUESTION [10 upvotes]: $\DeclareMathOperator{\tr}{tr}$Let's consider a $n \times n$ matrix and the sequence of traces $\tr(A)=\tr(A^{2})= \ldots = \tr(A^{n})=0$. How to prove that $A$ is a nilpotent matrix (a matrix so that $A^{k} \times u = 0$ for all $u \in V$ and for some $k$)?
Maybe it would be reasonable to consider $A$'s Jordan form (we assume that the action occurs over an algebraically closed field, such as $\mathbb{C}$). For example, suppose that we have found it and $A'$ consists of $m$ blocks of sizes $r_{1}, \ldots r_{m}$ ($r_{1}+r_{2}+\ldots+r_{m}=m, r_{i} \in \mathbb{Z}_{+}$) with $\alpha_{1}, \ldots \alpha_{m}$ as corresponding eigenvalues to each of the blocks. Then, according to the problem will get the system 
$$\begin{align}
r_{1}\alpha_{1}+&\ldots+r_{m}\alpha_{m}=0 \\
r_{1}{\alpha_{1}}^{2}+&\ldots+r_{m} {\alpha_{m}} ^{2}=0\\
&\ ~\,\vdots \\
r_{1}{\alpha_{1}}^{n}+&\ldots +r_{m}{\alpha_{m}}^{n}=0 
\end{align}$$
But the system's analysis over $\mathbb{C}$ seems to be too much complicated.
Actually, the problem must have more or less simple solution.
Any piece of advice would be much appreciated.

REPLY [3 votes]: To complete your first attempt of proof, you need the following lemma for Vandermonde matrices
Lemma: Let $V=\begin{bmatrix}
1 & \alpha_1 & \alpha_1^2 & \dots & \alpha_1^{n-1}\\
1 & \alpha_2 & \alpha_2^2 & \dots & \alpha_2^{n-1}\\
1 & \alpha_3 & \alpha_3^2 & \dots & \alpha_3^{n-1}\\
\vdots & \vdots & \vdots & \ddots &\vdots \\
1 & \alpha_n & \alpha_n^2 & \dots & \alpha_n^{n-1}
\end{bmatrix}$
Then $\det V=\prod\limits_{1\leq i<j\leq n}(\alpha_i-\alpha_j)$.
Corollary: $\det V=0\Leftrightarrow\exists i\neq j\ \alpha_i=\alpha_j$
Now, take the Jordan form as you suggested. Note that, for the purpose of calculating $tr A$ you need only non-zero eigenvalues.
Let $\alpha_1,\ldots,\alpha_m$ these eigenvalues $(m\leq n)$, $r_1,\ldots,r_m$ the dimensions of their generalized eigenspaces. You have the same relation as before
$r_{1}\alpha_{1}+\ldots+r_{m}\alpha_{m}=0 \\
r_{1}\alpha_{1}^{2}+\ldots+r_{m} \alpha_{m}^{2}=0\\
\ldots \\
r_{1}\alpha_{1}^{n}+\ldots +r_{m}\alpha_{m}^{n}=0$
That can be written
$$\begin{bmatrix}
\alpha_1 & \alpha_2 & \dots & \alpha_m\\
\alpha^2_1 & \alpha_2^2 & \dots & \alpha_m^2\\
\alpha_1^3 & \alpha_2^3 & \dots & \alpha_m^3\\
\vdots & \vdots & \ddots &\vdots \\
\alpha_1^n & \alpha_2^n & \dots & \alpha_m^n
\end{bmatrix}\begin{bmatrix} r_1\\r_2\\\vdots\\r_m\end{bmatrix}=0$$
Now, by hypothesis $\alpha_i$ are non-zero and differ one from another.
Now, take the first $m\times m$ minor of the system's coefficients matrix.
It's the transponse of a Vandermonde matrix V, each column of which has been multiplied by a non-zero element. The first lemma ensures that $\det V\neq 0$.
Therefore, the whole coefficient matrix has rank $m$, i.e. it represents an injective map $\mathbb{C}^m\rightarrow\mathbb{C}^n$. This implies $\forall i,\ r_i=0$, which tells you that the inital $A$ is indeed nilpotent.