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Aug 4, 2019 at 8:31 history edited David Roberts CC BY-SA 4.0
added 181 characters in body
Aug 4, 2019 at 8:26 comment added David Roberts $\log^2 n=(\log n)^2$? Or $\log(\log n)$? I'm guessing the former, but I initially thought the latter on reading it.
Aug 4, 2019 at 8:05 vote accept Chain Markov
S Aug 4, 2019 at 7:50 history suggested CommunityBot CC BY-SA 4.0
The original version (with $n = 2^m$) confused $2^{c m^3} = n^{B \log^2 n}$ with $2^{c m^2} = n^{B \log n}$. With the corrected exponent (as pointed out by Will Sawin) this response now gives a complete answer to the question.
Aug 4, 2019 at 3:35 review Suggested edits
S Aug 4, 2019 at 7:50
Aug 4, 2019 at 3:03 comment added Will Sawin There's a cube in the exponent for $p$-groups, so in fact we win by a mile.
Aug 4, 2019 at 2:20 review First posts
Aug 4, 2019 at 2:32
Aug 4, 2019 at 2:16 history answered Z3T3t3pON7 CC BY-SA 4.0