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I have to show that

$$\sqrt[7]{7!} < \sqrt[8]{8!}$$

and I did the following steps

\begin{align} \sqrt[7]{7!} &< \sqrt[8]{8!} \\ (7!)^{(1/7)} &< (8!)^{(1/8)} \\ (7!)^{(1/7)} - (8!)^{(1/8)} &< 0 \\ (7!)^{(8/56)} - (8!)^{7/56} &< 0 \\ (8!)^{7/56} \left(\left( \frac{7!}{8!} \right)^{(1/56)} - 1\right) &< 0 \\ \left(\frac{7!}{8!}\right)^{(1/56)} - 1 &< 0 \\ \left(\frac{7!}{8!}\right)^{(1/56)} &< 1 \\ \left(\left(\frac{7!}{8!}\right)^{(1/56)}\right)^{56} &< 1^{56} \\ \frac{7!}{8!} < 1 \\ \frac{1}{8} < 1 \\ \end{align}

Did I do this properly? Is this way the best way or is there another much easier way?

Thanks a bunch!

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1 Answer 1

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Your proof is correct, but perhaps a bit elaborate.

How about:

\begin{align} &&\sqrt[7]{7!} &< \sqrt[8]{8!}\\ \iff&&(7!)^8 &< (8!)^7\\ \iff&&(7!)^7 7! &< (8\cdot 7!)^7\\ \iff&&7! &< 8^7 \end{align}

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    $\begingroup$ Beat me to it. Well done. $\endgroup$
    – Pieter Geerkens
    Commented Nov 9, 2013 at 22:01
  • $\begingroup$ I'm slightly confused about your second step. How did you arrive at that? Did you just raise both sides to the power of the other term? $\endgroup$
    – Jeel Shah
    Commented Nov 9, 2013 at 23:23
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    $\begingroup$ @gekkostate I raised both sides to the $56$th power (which amounts to what you wrote, as $7$ and $8$ are coprime). $\endgroup$
    – Lord_Farin
    Commented Nov 9, 2013 at 23:32
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    $\begingroup$ @gekkostate Just remember the following rule for working with roots: try to get rid of them. :) It works surprisingly often. $\endgroup$
    – Lord_Farin
    Commented Nov 9, 2013 at 23:38
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    $\begingroup$ Is his proof correct? Isn't there a mistake when he takes the 8! to the power of thing outside of the thing? Certainly the fact that you have 7!<8^7 and he gets 7!<8! makes me think there is a problem somewhere... Though of course this does demonstrate why getting rid of the roots is such a good idea because they confuse things. :) $\endgroup$
    – Chris
    Commented Nov 10, 2013 at 1:29

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