order of the commutator subgroup divides $n^{n\left \lfloor \log_2 n \right \rfloor-n+2}$. Let $G$ be a group and $Z(G)$ be its center.
How can I prove the following:
If $(G:Z(G)) = n$ then $|G'|$ divides $n^{n\left \lfloor \log_2 n \right \rfloor-n+2}$.
I can prove that $(G')^n = 1$ and $|G'|<\infty$...

Since $|G/Z(G)| = n$, it can be generated by $d$ elements $g_1Z(G),\ldots,g_dZ(G)$, where $d \le \log_2 n$.
Let $H = \langle g_1,\ldots,g_d \rangle$ be the subgroup of $G$ generated by $g_1,\ldots,g_d$. Then $G' \le H$ and $|H/Z(G) \cap H| = |HZ(G)/Z(G)| = |G/Z(G)| = n$. So, by Schreier's Theorem, $Z(G) \cap H$ can be generated by at most $nd-n+1$ elements.
Hence $Z(G) \cap H \cap G' = Z(G) \cap G'$ can be generated by at most $nd-n+1$ elements, and, since $(G')^n=1$, $Z(G) \cap G'$ is a quotient of the direct product of $nd-n+1$ copies of a cyclic group of order $n$, and hence $|Z(G) \cap G'|$ divides $n^{nd-n+1}$.
The result now follows since $|G'/Z(G) \cap G'| = |G'Z(G)/Z(G)|$ divides $n$.
