Nested Radicals of Geometric Sequences

$\large \mathfrak{Q}=\sqrt[m]{1\sqrt[m]{n\sqrt[m]{n^2\sqrt[m]{n^3\cdots}}}}$ $\large =(n^{\frac{1}{m^2}})(n^{2(\frac{1}{m^3})})(n^{3(\frac{1}{m^4})})\cdots$ $\large =n^{\color{#20A900}{\frac{1}{m^2}+\frac{2}{m^3}+\frac{3}{m^4}+\frac{4}{m^5}\cdots}}$ Let $\large\mathfrak{S}=\frac{1}{m^2}+\frac{2}{m^3}+\frac{3}{m^4}+\frac{4}{m^5}\cdots (\small{\color{#302B94}{AGP}})~~$ $\dfrac{\mathfrak S}{m}=~~~~~~~~~~~~\frac{1}{m^3}+\frac{2}{m^4}+\frac{3}{m^5}+\cdots~~~~~~~~~$ Subtracting we get $\mathfrak S(1-\frac{1}{m})=\color{#D61F06}{\frac{1}{m^2}+\frac{1}{m^3}+\frac{1}{m^4}\cdots} (\small{\color{#302B94}{\text{Infinite GP}}})$
(For positive integer 'm' this is a Infinite GP) $\large \Rightarrow \mathfrak S=\frac{1}{(m-1)^2}$ $\large\Rightarrow \mathfrak Q=n^{\frac{1}{(m-1)^2}}$ For n=3, m=5 we get $\large \mathfrak Q=3^{\frac{1}{16}}= \sqrt[5]{1\sqrt[5]{3\sqrt[5]{9\sqrt[5]{27\cdots}}}}$ $\huge \therefore~16+3=\boxed{\color{#007fff}{19}}$

$3^{1/25}*9^{1/125}*27^{1/625}...=3^{1/25+2/125+3/625+...}=3^{1/16}=\sqrt[16]{3}$ so that the answer is $\boxed{19}$

The general case is $n^{1/m^2+2/m^3+3/m^4+...}=n^{1/(m-1)^2}$