This question already has an answer here:

I've been trying to work on a proof for this for a few days now, but can't seem to see the answer. It might be trivial but I would be grateful if anyone could let me know their strategy (if not a proof) for this:

$$\sqrt[n]{n!} \lt \sqrt[n+1]{(n+1)!}$$

Thanks in advance!


marked as duplicate by Martin R, mechanodroid, Claude Leibovici, Guy Fsone, Moishe Kohan Oct 30 '17 at 17:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Raise both side to the power of $n$ and then riase it to the power of $n+1$ and we have \begin{eqnarray*} (n!)^{n+1} < ((n+1)!)^n. \end{eqnarray*} Now cancel $(n!)^n$ \begin{eqnarray*} n! < (n+1)^n. \end{eqnarray*} This is obvious, multiply the following inequalities $1 <n+1,2<n+1, \cdots , n<n+1$.

  • $\begingroup$ Ah, of course, simplifying it makes it so obvious! Thank you so much! :) $\endgroup$ – mathnoob Oct 30 '17 at 0:44

Not the answer you're looking for? Browse other questions tagged or ask your own question.