Bariz - Problem

Given $T=\frac{2}{\sqrt{4-3\sqrt[4]{5} + 2\sqrt[4]{25}-\sqrt[4]{125}}}$. Let $S=4-3\sqrt[4]{5} + 2\sqrt[4]{5^{2}}-\sqrt[4]{5^{3}}$, we calculate $S \cdot \sqrt[4]{5}=4\sqrt[4]{5}-3\sqrt[4]{5^{2}} + 2\sqrt[4]{5^{3}}-5$. Adding $S+S \cdot \sqrt[4]{5}=-1+\sqrt[4]{5} -\sqrt[4]{5^{2}}+\sqrt[4]{5^{3}}$. We calculate $(S+S \cdot \sqrt[4]{5})\sqrt[4]{5}=-\sqrt[4]{5}+\sqrt[4]{5^{2}} -\sqrt[4]{5^{3}}+5$, Adding the previous result $S(1+\sqrt[4]{5})^{2}=4$ and $S=\frac{4}{(1+\sqrt[4]{5})^{2}}$. Then $T=\frac{2}{\sqrt{\frac{4}{(1+\sqrt[4]{5})^{2}}}}=1+\sqrt[4]{5}$. The general case has result $1+\sqrt[2n]{2n+1}$.