Power compression in the last lower-central term

Induct on $c$ for all finitely generated groups at once. For $c=1$, repeating a fixed word for $z$ gives $\mathrm{\mid }{z}^m{\mathrm{\mid }}_S\le \mathrm{\mid }z{\mathrm{\mid }}_S\mathrm{\mid }m\mathrm{\mid }$. The identity has length zero. Assume $c\ge 2$. The group $H={\gamma }_c$ is central, and by F1 it is generated by finitely many $t=[s,u]$ with $s\in S$, $u\in {\gamma }_{c-1}$, allowing inverses. It suffices first to bound each such t.

For $m\ge 1$ put $q=\lceil m^{1\mathrm{/}c}\rceil$ and divide $m=a{q}^{c-1}+b$ with $0\le b<q^{c-1}$. Then $0\le a\le q$ and $q\le 2m^{1\mathrm{/}c}$. In $G\mathrm{/}H$, the element $uH$ is in its last possible layer ${\gamma }_{c-1}(G\mathrm{/}H)$. If that quotient has smaller class, $uH=1$ and take empty words. Otherwise the induction hypothesis gives words for $(uH{)}^{q^{c-1}}$ and $(uH{)}^b$ of length at most $Kq$ (take the empty word when $b=0$). Lift their letters to S-words $v,w$ of the same length. Then $u^{q^{c-1}}=v{h}_1$, $u^b=w{h}_2$ for $h_1,h_2\in H$.

Because $[G,{\gamma }_{c-1}]\subseteq H$ is central, F2 gives $t^m=[s^a,u^{q^{c-1}}][s,u^b]=[s^a,v][s,w]$: multiplying either second input by the central element $h_i$ changes no commutator. The displayed word has length at most $2(a+Kq)+2(1+Kq)\le (4+4K)q\le (8+8K)m^{1\mathrm{/}c}$. Negative powers have the same length by inversion.

For fixed $z\in H$ choose a finite expression $z={\prod }_{j=1}^t{t}_j^{e_j}$ in these generators. Centrality gives $z^m={\prod }_j{t}_j^{e_{j}m}$. Subadditivity and step 3.1 bound its length by ${\sum }_{e_j\mathrm{\ne }0}{C}_{t_j}\mathrm{\mid }{e}_j{\mathrm{\mid }}^{1\mathrm{/}c}\mathrm{\mid }m{\mathrm{\mid }}^{1\mathrm{/}c}$. This finite sum is a valid $C_z$; the empty sum handles $z=1$. Exponent zero is the identity by the power convention.