The implication: $x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$ $$x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$$
The graphs/range: $\quad y_1(x)=x+\frac{1}{x}, \quad y_7(x)=x^7+\frac{1}{x^7} \quad$
and do not touch the line $\quad y=1\quad$.  The relation $\quad x+\frac 1 x=1 \quad$ appears to only be true for certain complex values
Yet the reasoning for the implication itself is
\begin{align}
x+\frac {1}{x}&=1 \\
\color {green}{x^2}&=\color {green}{x-1 }\\
x^3&=\color {green}{x^2}-x \\
&=(\color {green}{x-1})-x \\
&=-1.
\end{align}
Hence $x^6=1$, which implies $x^7=x$.
I had just started reading this book when it was presented, asking you to prove the implication without finding the roots first.
So my question: is this implication really justified?   Should the author have said to only prove it for an implicit (complex) specific value of $x$?  If I didn't know any better, I'd say he presented it like it was true for some range of values...he couldn't of...

You shouldn't be trying to find where $y_1=y_2$, but whether a root of $y_1$ is necessarily a root of $y_2$.
That is,
$$\forall x\Big(y_1(x)=0\implies y_2(x)=0\Big)$$
or equivalently,
$$\forall x\Big(y_1(x)\neq0\lor y_2(x)=0\Big)$$
