It is an old result of Schützenberger that in a free group, a basic commutator cannot be a proper power. A look at the original reference


M.-P. Schützenberger, *Sur l'équation $a^{2+n} = b^{2+m}c^{2+p}$ dans un groupe libre*, C. R. Acad. Sci. Paris 248 (1959), 2435–2436 (French).


quickly reveals that a lot of details are missing and some claims appear to be wrong.



> 
> **Question:** Let $F$ be a free group and $a,b,c \in F$ with $c \neq 1$ and $n \geq 2$. Why is $[a,b] \neq c^n$?
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In particular, it would be nice to have a somewhat geometric proof of this apparently fundamental fact. There is an algebraic proof in 


G. Baumslag, *Some aspects of groups with unique roots*, Acta Math 104(3) (1960), 217–303.


as Lemma 36.4 but it is relies on various technical computations and is hard to grasp.



A result of Schützenberger on commutators and powers in free groups
The case $n=2$ (originally due to Lyndon) admits a very nice geometric argument: one notes that elements $a,b,c$ with $[a,b]=c^2$ lead to a map from the surface $\Sigma\_{-1}$ of Euler characteristic -1 to a graph. Pulling back midpoints of edges, one obtains essential, two-sided, simple closed curves on $\Sigma\_{-1}$. Pinching these, Euler characteristic shows that one of the resulting components must be a projective plane, and the result follows.


For $n> 2$, [Duncan--Howie](http://link.springer.com/article/10.1007%2FBF02571522) proved the stronger theorem that, for any non-trivial element $c$ of $F\_2$, the stable commutator length of $c$ is at least $1/2$. (In fact, by passing to covers, one can also deduce the $n=2$ case from the Duncan--Howie theorem.) When you decode it, their proof is quite geometric, relying on defining a sort of combinatorial vector field on surfaces.


[Louder and I](https://arxiv.org/abs/1410.2540) gave a sort of generalization of Duncan--Howie's proof in our work on one-relator groups. Again, the idea is to count Euler characteristic using some sort of geometric data (we call it a *stacking*). We don't spell it out explicitly, but you can use our ideas to give a geometric proof of the Duncan--Howie theorem, and in particular of the result that you want.


