Let $A$ be the free associative algebra over a field $k$ generated by countably many indeterminates $x_1, x_2, \ldots$.
I want to show that for any $n$, $x_1 \ldots x_n$ is not in the ideal $I$ generated by $S=\{x_i^2, x_ix_j+x_jx_i : i,j \geq 0\}$.
My attempt: Suppose for a contradiction that it is. Then it would be a sum of terms of the form $xsy$ where $x,y \in A$ and $s \in S$. I would like to argue that each such term $xsy$ must be $0$ since $s$ has the form $x_i^2$ (so it has degree $2$ in $x_i$, whereas $x_1 \ldots x_n$ does not) or $x_ix_j+x_jx_i$ (and this is 'bad' because it contributes two terms where the orders in which $x_i$ and $x_j$ occur are flipped). A problem with this argument (other than it being too informal), is that it might be that (the 'bad' part of) such a term $xsy$ cancels with (the 'bad' part of) another term $xs'y$, neither term being $0$. How do I correct and formalize my argument?
Footnote: I am only interested in the case where $k$ has characteristic not equal to $2$; in this case the ideal $I$ can be generated just by the relations $\{ x_ix_j+x_jx_i : i,j \geq 0\}$, since $x_i^2 = \frac{1}{2} (x_ix_i+x_ix_i)$. Feel free to assume this, but if there is a proof that works in full generality even in characteristic $2$ I would love to see it.