Trace of a nilpotent matrix is zero. [duplicate]

I'm trying to prove that the trace of a nilpotent matrix is $0$. Here's what I have:

Let $A \in M_n(k)$, where $k$ is a field. If $A$ is nilpotent then, in fact, $A^n=0$, (this result was proven in the previous problem.) By Proposition 20 on page 428 (Dummit & Foote) the characteristic polynomial of $A$ divides some power of the minimal polynomial of $A$, and so is of the form $x^m$ for some $m \ge t$. Thus, the eigenvalues of $A$ are all equal to $0$ and the only solution to the equation $xv-Av=0$ is the trivial solution. Since it is assumed that $v \neq 0$, then it must be that $A=0$.

This is a much stronger conclusion than the conclusion I'm aiming for and so I am suspicious. I probably made an error. Please help.

To expand on the comment by Pedro Tamaroff, I can add the following.

The initial part of your argument seems to aim at the (correct) conclusion that the characteristic polynomial must be $X^n$ (remember that unlike the minimal polynomial, you know the degree of the characteristic polynomial beforehand). But you never really make that conclusion, which you could reach as follows: since some $X^k$ annihilates $A$, the minimal polynomial divides $X^k$ and must be a power of $X$, but then the characteristic polynomial, which divides a power of the minimal polynomial, must also be a power of $X$, necessarily $X^n$. (Note that this starts with the order $k$ of nilpotency of$~A$; there is no need to argue for $k=n$ initially, although the conclusion of the argument shows, by Cayley-Hamilton, that $X^n$ annihilates $A$ as well.)

Once you know the characteristic polynomial, you can simply read off the trace as minus its subdominant coefficient, which is zero for $X^n$.

A more low-tech argument is possible as well, provided you know that the trace is invariant under change of basis. The chain of subspaces $\{0\}=\ker A^0\subseteq\ker A^1\subseteq\ker A^2\subseteq\cdots$ grows up to the whole space for nilpotent$~A$. Choosing a basis of $\ker A^1$, then extending to a basis of the next space $\ker A^2$, and so on, eventually gives a basis of the whole space. By construction change of basis of $A$ to this new basis leads to a strictly upper triangular matrix; its trace is obviously$~0$.

Even if $A\ne 0$ has all of its eigenvalues $0$ it doesn't mean $Av=xv$ has no nontrivial solutions; consider for instance $A=(\begin{smallmatrix}0&1\\0&0\end{smallmatrix})$ and $v=(\begin{smallmatrix}1\\0\end{smallmatrix})$.

Anyway, to do the problem, know/use the fact that the trace is the sum of eigenvalues.