# What is sqrt(7sqrt(7sqrt7)) ?

What is $\sqrt{7\sqrt{7\sqrt{7}}}$ ?
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Lets asume:
$y=\sqrt{7\sqrt{7\sqrt{7}}}$,
${y}^{2}=7\sqrt{7\sqrt{7}}$,
${y}^{4}={7}^{2}\cdot 7\sqrt{7}={7}^{3}\sqrt{7}$,
${y}^{8}={7}^{6}\cdot 7={7}^{7}$
So $y={7}^{\frac{7}{8}}=5.4886$ . In general, if N is the number of 7's under the square root then $y={7}^{\frac{\left({2}^{N}\right)-1}{{2}^{N}}}$. In this case N=3 so $y={7}^{\frac{{2}^{3}-1}{{2}^{3}}}={7}^{\frac{7}{8}}$. For N=4, $y={7}^{\frac{15}{16}}=6.1984$, as N approaches infinity y approaches 7. So for 7's repeated under square roots indefinitely the answer is 7.
Another way of solving it is to put $y=\sqrt{7\left(\sqrt{7}\left(\sqrt{7}\dots }$ So ${y}^{2}=7\sqrt{7}\left(\sqrt{7}\sqrt{7}\dots =7y$, so since $y\ne 0,y=7$ (dividing through by y).