Commutator with a generator of a free group Let $F$ be a free group $\langle x_1,...x_m\rangle$.
If $a\in F_2$ and $[a,x_1] \in F_n$ then $a\in F_{n-1}$.
Here, $F_n$ is the $n$-th lower central series term with $F_2=[F:F]$.
How can I prove this fact?
 A: (Edited)
There may be a clever geometric argument to be made using the fact that free groups are fundamental groups of bouquets of circles, but I don't know it. I would do this using commutator calculus. If I'm not mistaken, you can essentially follow the arguments I used in the paper: 

Capability of nilpotent products of cyclic groups II, Journal of Group Theory 10 (2007), no. 4, 441-451 (MR 2334753)

to compute the center of nilpotent products of cyclic groups in the small class case.
You can find some of the basic stuff on basic commutators in Marshal Hall's book; an encyclopedic source is:

Ward, M.A. Basic Commutators, Philos. Trans. Roy. Soc. London Series A, 264 (1969), 343-412, MR 40:#4379.

Let me give you a quick rundown. The basic concept is that of basic commutators: the basic commutators on $x_1,\ldots,x_n$ and their weights are defined recursively as follows:


*

*$x_1,\ldots,x_n$ are basic commutators of weight $1$, ordered by $x_i\lt x_j$ if and only if $i\lt j$.

*Assuming all basic commutators of weight less than $k$ have been defined an ordered, the basic commutators of weight exactly $k$ are commutators of the form $[c_j,c_i]$, where $c_i$ and $c_j$ are basic commutators, $\mathrm{wt}(c_i)+\mathrm{wt}(c_j)=k$, and $c_i<c_j$; in addition, if $c_j = [c_r,c_s]$, with $c_r$ and $c_s$ basic commutators, then $c_s\leq c_i$. 

*Order the basic commutators of weight exactly $k$ amongst themselves lexicographically right to left, so $[c_r,c_s] \lt [c_j,c_i]$ if and only if $c_s\lt c_i$, or $c_s=c_i$ and $c_r\lt c_j$; and making them larger than those of weight less than $k$.


For example, the first few basic commutators on $x_1,x_2,x_3$ are, in order,
$$x_1,\ x_2,\ x_3,\ [x_2,x_1],\ [x_3,x_1],\ [x_3,x_2],\ [x_2,x_1,x_1],\ [x_3,x_1,x_1],\ [x_2,x_1,x_2],\ [x_3,x_1,x_2],\ [x_3,x_2,x_2],\ [x_2,x_1,x_3],\ [x_3,x_1,x_3],\ [x_3,x_2,x_3],\ldots$$
where commutators are left-normed, so $[a,b,c]=[[a,b],c]$. Once you get to weight $4$, you'll have both commutators of the form $[a,b,c,d]$, and also some of the form $[[a,b],[c,d]]$, etc.
For any $n$, the elements of $F/F_{n+1}$ can be written uniquely as
$$x_1^{a_1}\cdots x_n^{a_n}c_1^{\alpha_1}\cdots c_m^{\alpha_m}$$
where $c_1\lt c_2\lt \cdots \lt c_m$ are the basic commutators of weights at least two and at most $n$, and both the $a_i$ and $\alpha_j$ are integers. This is achieved through the collection process. In particular, the free abelian group $F_k/F_{k+1}$ has basis given by the basic commutators of weight exactly $k$.
There are many known formulas for expressing certain commutators in terms of basic commutators. The key concept you'll need for the calculation you want is that of the commutator $[u\leftarrow v]$ (I don't know of an official pronunciation for it):
Given basic commutators $u$ and $v$, define the basic commutator $[u\leftarrow v]$ as follows:


*

*If $v\gt u$, then $[u\leftarrow v] = [v\leftarrow u]$.

*If $v\lt u$ and $u=[c_1,c_2]$ and $c_2\gt v$, then $[u\leftarrow v] = [c_1\leftarrow v,c_2]$.

*Otherwise, $[u\leftarrow v] = [u,v]$.


We have that $[u\leftarrow v]$ is always a basic commutator of weight $\mathrm{wt}(u)+\mathrm{wt}(v)$. And we have the following result:

Proposition. Let $F$ be the free group on $x_1,\ldots,x_n$, and let $u$ and $v$ be basic commutators, with $k=\mathrm{wt}(u)+\mathrm{wt}(v)$. Then
  $$ [u,v] \equiv [u\leftarrow v]^{\epsilon}c_1^{\alpha_1}c_2^{\alpha_2}\cdots c_m^{\alpha_m} \pmod{F_{k+1}}$$
  where $\epsilon=\pm 1$, $\alpha_i$ are integers, each $c_i$ is a basic commutator of weight $k$, and $[u\leftarrow v]\lt c_1\lt\cdots \lt c_m$. Moreover, if $u\geq v$, we may take $\epsilon = 1$. 

This result will give  you essentially what you want, once you have some of the basic properties of $[u\leftarrow v]$ (for example, if $v\lt u_1\lt u_2$, then $[u_1\leftarrow v]\lt [u_2\leftarrow v]$; and if $u\gt v_1\gt v_2$, then $[u\leftarrow v_1]\gt[u\leftarrow v_2]$). 
Take an element $a\in F_2$, and write it as $a = d_1^{a_1}\cdots d_m^{a_m}c$, with $d_1\lt d_2\lt\cdots \lt d_m$ basic commutators of weight at least $2$ and at most $n-1$, and $c\in F_n$. Then the "leading term" of $[a,x_1]$ modulo $F_{n}$ will be $[d_1\leftarrow x_1]^{a_1}$ (you'll need some commutator calculus to show this). If this lies in $F_n$, then the weight of $[d_1\leftarrow x_1]$ must be at least $n$, which means the weight of $d_1$ must be at least $n-1$, and hence $a$ is a product of basic commutators of weight at least $n-1$, hence $a\in F_{n-1}$, as desired.
