Let $r$ be a natural number. The elements of the free group $F_r$ on $r$ generators have a nice concrete description as "words" in the $r$ generators (and their inverses). I'd like to know if there is anything analogous for its profinite completion.

Namely, let $\hat{F}_r$ be the free profinite group on $r$ generators. Then $\hat{F}_r$ contains the free group $F_r$ as a dense subgroup, and every element of $\hat{F}_r$ can be written as an infinite product $\prod_{i=1}^\infty x_i$ where each $x_i \in F_r$ and the $x_i$ tend to $1$ in the profinite topology.

In other words, we can arrange that $x_i \to 1$ in the following sense: for every $N$, the $x_i$ have the property that almost all of them are mapped to $1$ by every homomorphism $F_r \to G$ where $G$ is any group of cardinality $\leq N$ (for $i \gg 0$, depending on $N$). This suggests a natural descending filtration on $F_r$ via the intersection of the kernels of all maps $F_r \to G$ where $|G|\leq N$. Given a nice description (e.g., generators) of the elements in this filtration, one would have a somewhat explicit procedure of describing elements in the profinite completion via "infinite" words.

There are some natural elements in filtration $N$ in $F_r$, namely the $c(N)$th powers where $c(N)$ is the least common multiple of $\{1, 2, \dots, N\}$.

**Question:** Is the $N$th filtration in $F_r$ generated by the $c(N)$th powers? That is, if $a \in F_r$ is an element that goes to $1$ in every finite group of cardinality $\leq N$ a (finite) product of $c(N)$th powers?

**Question${}^\prime$:** If the above is too much to expect, are the two possible filtrations on $F_r$ above at least "commensurable"?