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S Jan 29, 2018 at 10:02 history suggested Jose Brox CC BY-SA 3.0
Typo in formula
Jan 29, 2018 at 9:27 review Suggested edits
S Jan 29, 2018 at 10:02
Jan 29, 2018 at 9:26 comment added Jose Brox Thanks, I thought that perhaps you had a different approach. I'll do a little edit.
Jan 29, 2018 at 7:59 comment added abx You are right, I made a mistake in the computation. That doesn't affect the conclusion.
Jan 28, 2018 at 19:00 comment added Jose Brox What do you exactly do to see $v_p(d!)\geq \frac{d-1}{p-1}-\log_p(d)$? If we use $[\frac{d}{p^i}]\geq \frac{d}{p^i}-1$ then, since the sum goes up to $[\log_p(d)]$, we get $d-dp^{-[\log_p(d)]}\leq d-1$ in the numerator.
Jan 21, 2018 at 15:33 history answered abx CC BY-SA 3.0