Let$$x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}.$$Find $(x+1)^{48}$.

It is   done like this  

\(x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}.\\~\\ find \;(x+1)^{48}\\~\\ consider (\sqrt5+1)(\sqrt[4]{5}+1)(\root 8\of5+1)(\root {16}\of5+1)\\ let\;\; y=\sqrt[16]{5}\\ y^2=\sqrt[8]{5}\\ y^4=\sqrt[4]{5}\\ y^8=\sqrt[2]{5}\\ y^{16}=5\\ \text{so I now have}\\ (y^8+1)(y^4+1)(y^2+1)(y+1)\\ \text{when expanded out I get}\\ y^{15}+y^{14}+y^{13}+......+y^1+1\\ \text{This is the sum of a GP}\\ sum=\frac{y^{16}-1}{y-1}\\~\\ x+1=\frac{4}{the \;sum}+1\\ x+1=4\div \frac{y^{16}-1}{y-1}+1\\ x+1= \frac{4(y-1)}{y^{16}-1}+1\\ x+1= \frac{4(\sqrt[16]{5}-1)}{5-1}+1\\ x+1=\sqrt[16]{5}\\ (x+1)^{48}=(\sqrt[16]{5})^{48}\\ (x+1)^{48}=[(\sqrt[16]{5})]^{16}]^3\\ (x+1)^{48}=5^3\\ (x+1)^{48}=125\)

LaTex coding:

x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}.\\~\\
find \;(x+1)^{48}\\~\\
consider
(\sqrt5+1)(\sqrt[4]{5}+1)(\root 8\of5+1)(\root {16}\of5+1)\\
let\;\; y=\sqrt[16]{5}\\
y^2=\sqrt[8]{5}\\
y^4=\sqrt[4]{5}\\
y^8=\sqrt[2]{5}\\
y^{16}=5\\
\text{so I now have}\\
(y^8+1)(y^4+1)(y^2+1)(y+1)\\
\text{when expanded out I get}\\
y^{15}+y^{14}+y^{13}+......+y^1+1\\
\text{This is the sum of a GP}\\
sum=\frac{y^{16}-1}{y-1}\\~\\
x+1=\frac{4}{the \;sum}+1\\
x+1=4\div \frac{y^{16}-1}{y-1}+1\\
x+1= \frac{4(y-1)}{y^{16}-1}+1\\
x+1= \frac{4(\sqrt[16]{5}-1)}{5-1}+1\\
x+1=\sqrt[16]{5}\\
(x+1)^{48}=(\sqrt[16]{5})^{48}\\
(x+1)^{48}=[(\sqrt[16]{5})]^{16}]^3\\
(x+1)^{48}=5^3\\
(x+1)^{48}=125

Melody did get it right :) But I'd like to point out an easier method.

Multiply by \(\frac{\sqrt[16]{5}-1}{\sqrt[16]{5}-1}\), and the expression simplifies to \(\sqrt[16]{5}-1\). Our answer is hence \((\sqrt[16]{5})^{48} = \boxed{125}\).