# Is my proof correct for: $\sqrt[7]{7!} < \sqrt[8]{8!}$

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!

• @gekkostate I raised both sides to the $56$th power (which amounts to what you wrote, as $7$ and $8$ are coprime). – Lord_Farin Nov 9 '13 at 23:32