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. 

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 $n!A^n=0$.
