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May 18, 2018 at 21:10 history edited Michael Albanese CC BY-SA 4.0
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Nov 3, 2016 at 23:40 history edited Alexey Ustinov CC BY-SA 3.0
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Jun 5, 2012 at 12:05 vote accept Igor Belegradek
Jun 4, 2012 at 23:24 comment added Igor Belegradek Thank you. One needs also a little combinatorial argument on how many powers of two are in $(2k)!$, but I think I got it.
Jun 4, 2012 at 21:27 history answered Igor Rivin CC BY-SA 3.0