Let $R$ be a commutative ring with unity such that $n!$ is not a zero-divisor. Let $s_1=\sigma_1,s_2,s_3\cdots$ and $\sigma_1,\sigma_2\cdots$, (convention: if $k>n$, then $\sigma_k=0$) be elements of $R$. We consider the Newton's identities

$s_k=\sigma_1s_{k-1}-\cdots+(-1)^k\sigma_{k-1}s_1+(-1)^kk\sigma_k$.

We assume that there is a zero sequence of length $n$: $s_p=\cdots=s_{p+n-1}=0$. Formal calculations seem to indicate that the $(\sigma_i)_{1\leq i\leq n}$ are in the nilradical of $R$. In particular, the result is true for $(n=6,p\leq9),(n=7,p\leq8),(n=8,p\leq 7),(n=9,p\leq5)$.

Is this result true in general ?

If that helps, in Section III.2 of my article http://link.springer.com/article/10.1007/s00023-003-0127-7 there is a proof due to de Calan and Magnen of the fact: for any $k\ge 1$, $({\rm tr}\ N)^{k(n-1)+1}$ is in the ideal generated by the matrix elements of $N^k$. Here $N$ is an $n\times n$ matrix of indeterminates.