$\operatorname{tr}(A^k)=0 \space \forall k\in \Bbb{Z}^+$ implies $A$ is nilpotent. Does this imply $\operatorname {char}(K) =0$? $\mathcal{M}_{n}(K) $: Set of all $n×n$  matrices over the field $K$.
$A\in \mathcal{M}_{n}(K) $ is called nilpotent if $A^k=\textbf{0}$ for some $k\in \Bbb{Z}^+$

It is clear that if $A$ is nilpotent then $\operatorname{tr}(A^k) =0$ for all $k\in \Bbb{Z}^+$

The converse is also true if $\operatorname{char}(K) =0$

My question: Suppose for a matrix $A\in\mathcal{M}_n(K)$, $\operatorname{tr}(A^k)=0 \space  \forall k\in \Bbb{Z}^+ $ implies $A$ is nilpotent.
Can we conclude that $\operatorname{char}(K) =0$?
 A: Let $P(n)$ be the statement "If $A \in M_n(K)$ is such that for all $k \ge 1$, $\operatorname{tr}(A^k)=0$, then $A$ is nilpotent."
Claim: $P(n) \Leftrightarrow (\operatorname{char}(K) = 0 \text{ or } n < \operatorname{char}(K) $
Proof of "$\Rightarrow$": Assume $n \ge \operatorname{char}(K) =: p$. Consider the matrix whose first $p$ diagonal entries are $1$, all other entries $0$.
Proof of "$\Leftarrow$": See e.g. the answer to Trace Criterion. Notice for the case of positive characteristic $p$ that under our assumption, $\operatorname{tr}(Id_{m\times m}) = m \neq 0 \in K$ for all $1 \le m \le n < p$, which makes the argument in that answer go through.
(It would be interesting to see which of the proofs in Traces of all positive powers of a matrix are zero implies it is nilpotent, $\operatorname{tr}(A)=\operatorname{tr}(A^{2})= \ldots = \operatorname{tr}(A^{n})=0$ implies $A$ is nilpotent, The characteristic polynomial of $A$ is $x^n$ if and only if $\text{Tr}(A^i)=0$ for all $1\le i \le n$. also go through assuming $n < p$.)

So of course
$$(\forall n: P(n)) \Leftrightarrow \operatorname{char}(K)=0$$
