(a) On the free 2-step nilpotent group $L$ on $(u,v)$, the commutator $z=[u,v]$ is not a proper power.

Indeed, since $L/[L,L]$ is torsion-free, any root of $z$ should lie in $[L,L]$ and the latter is readily seen to be the cyclic group generated by $z$. (Remark: $L$ is isomorphic to the integral 3-dimensional Heisenberg group.)$\qquad\Box$

I'll use three other very classical facts, without proof: the hardest (b) has both geometric and combinatorial proofs (the simplest using that $\pi_1$ of graphs are free), while (c),(d) are elementary exercises ((c) follows from Malcev's more general results but is very easy by hand here).

(b) Subgroups of free groups are free.

 

(c) $L$ is Hopfian (all its surjective group endomorphisms are automorphisms).

 

(d) If $N$ is a nilpotent group and $M$ a subgroup generating $N$ modulo $[N,N]$ (that is, $M[N,N]=N$), then $M=N$.

This allows to prove the result.

In a free group $F$, if $a,b,c$ are two elements and $[a,b]\neq 1$ then $[a,b]$ is not a proper power $c^n$ of $c$.

By (a), we can suppose that $F$ is generated by $a,b,c$. If $d$ is the rank of $F$, it is also the rank of its abelianization, and the relation $[a,b]=c^n$ then forces $d\le 2$. Also $d\ge 2$ since $a,b$ don't commute. So $F$ is free on 2 generators (up to now this is exactly Baumslag's argument).

Hence $L=F/[F,[F,F]]$ is free 2-step nilpotent on 2 generators. Denote by $A,B,C$ the images of $a,b,c$ in $L$. Since $L/[L,L]$ is torsion-free and $C^n\in [L,L]$, we have $C\in [L,L]$. By (d), we deduce that $L$ is generated by $A,B$. So the endomorphism of $L$ mapping the two free generators to $A$ and $B$ is surjective. By (c) deduce that $L$ is free 2-step nilpotent on $(A,B)$. Since $[A,B]=C^n$, we can use (a) to get a contradiction. (Baumslag's conclusion of the argument also uses a 2-step nilpotent group, but I haven't followed as it's more technical)

(a) On the free 2-step nilpotent group $L$ on $(u,v)$, the commutator $z=[u,v]$ is not a proper power.

Indeed, since $L/[L,L]$ is torsion-free, any root of $z$ should lie in $[L,L]$ and the latter is readily seen to be the cyclic group generated by $z$. (Remark: $L$ is isomorphic to the integral 3-dimensional Heisenberg group.)$\qquad\Box$

I'll use three other very classical facts, without proof: the hardest (b) has both geometric and combinatorial proofs (the simplest using that $\pi_1$ of graphs are free), while (c),(d) are elementary exercises ((c) follows from Malcev's more general results but is very easy by hand here).

(b) Subgroups of free groups are free.

(c) $L$ is Hopfian (all its surjective group endomorphisms are automorphisms).

(d) If $N$ is a nilpotent group and $M$ a subgroup generating $N$ modulo $[N,N]$ (that is, $M[N,N]=N$), then $M=N$.

This allows to prove the result.

In a free group $F$, if $a,b,c$ are two elements and $[a,b]\neq 1$ then $[a,b]$ is not a proper power $c^n$ of $c$.

By (a), we can suppose that $F$ is generated by $a,b,c$. If $d$ is the rank of $F$, it is also the rank of its abelianization, and the relation $[a,b]=c^n$ then forces $d\le 2$. Also $d\ge 2$ since $a,b$ don't commute. So $F$ is free on 2 generators (up to now this is exactly Baumslag's argument).

Hence $L=F/[F,[F,F]]$ is free 2-step nilpotent on 2 generators. Denote by $A,B,C$ the images of $a,b,c$ in $L$. Since $L/[L,L]$ is torsion-free and $C^n\in [L,L]$, we have $C\in [L,L]$. By (d), we deduce that $L$ is generated by $A,B$. So the endomorphism of $L$ mapping the two free generators to $A$ and $B$ is surjective. By (c) deduce that $L$ is free 2-step nilpotent on $(A,B)$. Since $[A,B]=C^n$, we can use (a) to get a contradiction. (Baumslag's conclusion of the argument also uses a 2-step nilpotent group, but I haven't followed as it's more technical)

(a) On the free 2-step nilpotent group $L$ on $(u,v)$, the commutator $z=[u,v]$ is not a proper power.

Indeed, since $L/[L,L]$ is torsion-free, any root of $z$ should lie in $[L,L]$ and the latter is readily seen to be the cyclic group generated by $z$. (Remark: $L$ is isomorphic to the integral 3-dimensional Heisenberg group.)$\qquad\Box$

I'll use three other very classical facts, without proof: the hardest (b) has both geometric and combinatorial proofs (the simplest using that $\pi_1$ of graphs are free), while (c),(d) are elementary exercises ((c) follows from Malcev's more general results but is very easy by hand here).

(b) Subgroups of free groups are free.

(c) $L$ is Hopfian (all its surjective group endomorphisms are automorphisms).

(d) If $N$ is a nilpotent group and $M$ a subgroup generating $N$ modulo $[N,N]$ (that is, $M[N,N]=N$), then $M=N$.

This allows to prove the result.

In a free group $F$, if $a,b,c$ are two elements and $[a,b]\neq 1$ then $[a,b]$ is not a proper power $c^n$ of $c$.

By (a), we can suppose that $F$ is generated by $a,b,c$. If $d$ is the rank of $F$, it is also the rank of its abelianization, and the relation $[a,b]=c^n$ then forces $d\le 2$. Also $d\ge 2$ since $a,b$ don't commute. So $F$ is free on 2 generators. Hence $L=F/[F,[F,F]]$ is free 2-step nilpotent on 2 generators. Denote by $A,B,C$ the images of $a,b,c$ in $L$. Since $L/[L,L]$ is torsion-free and $C^n\in [L,L]$, we have $C\in [L,L]$. By (d), we deduce that $L$ is generated by $A,B$. So the endomorphism of $L$ mapping the two free generators to $A$ and $B$ is surjective. By (c) deduce that $L$ is free 2-step nilpotent on $(A,B)$. Since $[A,B]=C^n$, we can use (a) to get a contradiction.

(a) On the free 2-step nilpotent group $L$ on $(u,v)$, the commutator $z=[u,v]$ is not a proper power.

Indeed, since $L/[L,L]$ is torsion-free, any root of $z$ should lie in $[L,L]$ and the latter is readily seen to be the cyclic group generated by $z$. (Remark: $L$ is isomorphic to the integral 3-dimensional Heisenberg group.)$\qquad\Box$

I'll use three other very classical facts, without proof: the hardest (b) has both geometric and combinatorial proofs (the simplest using that $\pi_1$ of graphs are free), while (c),(d) are elementary exercises ((c) follows from Malcev's more general results but is very easy by hand here).

(b) Subgroups of free groups are free.

(c) $L$ is Hopfian (all its surjective group endomorphisms are automorphisms).

(d) If $N$ is a nilpotent group and $M$ a subgroup generating $N$ modulo $[N,N]$ (that is, $M[N,N]=N$), then $M=N$.

This allows to prove the result.

In a free group $F$, if $a,b,c$ are two elements and $[a,b]\neq 1$ then $[a,b]$ is not a proper power $c^n$ of $c$.

By (a), we can suppose that $F$ is generated by $a,b,c$. If $d$ is the rank of $F$, it is also the rank of its abelianization, and the relation $[a,b]=c^n$ then forces $d\le 2$. Also $d\ge 2$ since $a,b$ don't commute. So $F$ is free on 2 generators (up to now this is exactly Baumslag's argument). 

Hence $L=F/[F,[F,F]]$ is free 2-step nilpotent on 2 generators. Denote by $A,B,C$ the images of $a,b,c$ in $L$. Since $L/[L,L]$ is torsion-free and $C^n\in [L,L]$, we have $C\in [L,L]$. By (d), we deduce that $L$ is generated by $A,B$. So the endomorphism of $L$ mapping the two free generators to $A$ and $B$ is surjective. By (c) deduce that $L$ is free 2-step nilpotent on $(A,B)$. Since $[A,B]=C^n$, we can use (a) to get a contradiction. (Baumslag's conclusion of the argument also uses a 2-step nilpotent group, but I haven't followed as it's more technical)