Overpowered $x$ 's - 1

Note that $\large {(x + \frac {1} {x})} ^ {2} = x^2 + \frac {1} {x^2} +2$ .

If $\large x+ \frac {1} {x} = 2$ then the equation $\large {x} ^ {{2} ^ {n}}+{(\frac {1} {x})} ^ {{2} ^ {n}}= 2$ is valid for all integer $n \geq 0$ . The problem asks for $n = 7$ , which is $2$ .

$\begin{aligned} x + \frac{1}{x} & = 2 \\ \Rightarrow x^2 - 2x + 1 & =0\\ (x-1)^2 & =0 \\ x & =1\\ \Rightarrow x^{128}+ \frac{1}{x^{128}}&= \boxed {2} \end{aligned}$

Let $y={ x }^{ 128 }$ . Then, ${ x }^{ 128 }+\frac { 1 }{ { x }^{ 128 } } =y+\frac { 1 }{ y } =2$ .

$x+\frac{1}{x} \geq 2$ by AM-GM with equality iff $x=1$ so we have that $x+\frac{1}{x}=2 \Rightarrow x=1 \Rightarrow x^{128}+ \frac{1}{x^{128}} =2$