Calculate the value of the radical: $$\left[\sqrt[{\sqrt[\sqrt3]{\sqrt3}}]{(3 \sqrt 3)^{ \sqrt[\sqrt3]{\sqrt3^{\sqrt3+1}}}}\right]^{\sqrt 3}$$
What is the tip to solve this problem? I think $(3\sqrt{3})^{9}$ is the answer
Attemp: I started thinking about replacing $ \sqrt {3} $ for $y$, but I can't develop much
Just do it. But do it slowly and patiently.
$\left[\sqrt[\color{green}{{\sqrt[\sqrt3]{\sqrt3}}}]{(3 \sqrt 3)^{ \color{red}{ \sqrt[\sqrt3]{\sqrt3^{\sqrt3+1}}}}}\right]^{\color{purple}{\sqrt 3}}=$
$(3\sqrt 3)^{\frac {\color{purple}{\sqrt 3}\color{red}{ \sqrt[\sqrt3]{\sqrt3^{\sqrt3+1}}}}{\color{green}{{\sqrt[\sqrt3]{\sqrt3}}}}}=$
$(3\sqrt 3)^{(\sqrt 3)^{\color{purple}{1}-\color{green}{\sqrt 3}}\color{red}{\sqrt{3}^{\frac{\sqrt 3+1}{\sqrt 3}}}}=$
$(3\sqrt 3)^{\sqrt{3}^{1-\sqrt 3+ \frac {\sqrt 3+1}{\sqrt 3}}}=$
$(3\sqrt 3)^{\sqrt{3}^{1-\sqrt 3+ 1+\sqrt 3}}=$
$(3\sqrt 3)^{\sqrt 3^2}=$
$(3\sqrt 3)^3 =$
$(3^13^{\frac 12})^3=$
$(3^{1+\frac 12})^3 =$
$(3^{\frac 32})^3=$
$3^{\frac 92}=$
$3^{4\frac 12}=$
$3^4\sqrt 3=$
$81\sqrt 3$