On $n\times n$ matrices $A$ with trace of the powers equal to $0$ 
Let $R$ be a commutative ring with identity and let $A \in M_n(R)$ be such that $$\mbox{tr}A = \mbox{tr}A^2 = \cdots = \mbox{tr}A^n = 0 .$$ I want to show that $n!A^n= 0$. 

Any suggestion or reference would be helpful.  
P.S.: When $R$ is a field of characteristic zero I can prove that $A^n=0$ but I have no idea for the general case. 
 A: Let 
$$\chi(\lambda) = \det(\lambda I_n - A ) = \lambda^n - e_1 \lambda^{n-1} + e_2 \lambda^{n-2} + \cdots + (-1)^n e_n$$
be the characteristic polynomial of $A$. By Cayley–Hamilton theorem (which is valid for any commutative ring), we have
$$\chi(A) = A^n - e_1 A^{n-1} + e_2 A^{n-2} + \cdots + (-1)^n e_n I_n = 0\tag{*1}$$
Let $p_k = \text{tr}A^k$ for $k = 1, \ldots, n$. The $e_k$ are elementary symmetric polynomials in the roots of $\chi(\lambda)$ and they are related to $p_k$ through the Newton identities (again valid for any commutative ring):
$$\begin{align}
e_1   = & p_1,\\
2 e_2 = & e_1 p_1 - p_2,\\
3 e_3 = & e_2 p_1 - e_1 p_2 + p_3,\\
4 e_4 = & e_3 p_1 - e_2 p_2 + e_1 p_3 - p_4,\\
\vdots
\end{align}$$
As a result, 
$$p_1 = p_2 = \cdots = p_n = 0\quad\implies\quad e_1 = 2 e_2 = 3 e_3 = \cdots = n e_n = 0$$
Multiply $(*1)$ by $n!$ then give us $\;n!A^n = 0\;$.
A: The following argument also works in prime characteristic. The coefficients
of the characteristic polynomial 
$$
\chi(t)=\sum^n_{j=0}
(-1)^j \omega_j (A)\:t^{n-j}\; 
$$ of $A$ satisfy the following identities:
$$
\sum^j_{i=1} (-1)^{i+1} {\rm tr}(A^i)\:\omega_{j-i} (A) =j\cdot \omega_j (A)
\hbox{ with }  \omega_0 (A)=1,\; \omega_{n+j}(A) = 0 \quad \forall \;
j\in \mathbb{N}
$$
We have either $p\mid n!$, or we have
$p>n$ so that $\omega_j (A) = 0 \quad\forall \, j \ge 1$. In this case $A$ has
characteristic polynomial $t^n$, so that $A$ is nilpotent with $A^n=0$. Together this means $n!A^n=0$.
