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...

The given equation implies $x^2-x+1=0$; multiplying by $x+1$, we get $x^3+1=0$. This implies $x^6=1$ and therefore
$$
x^7=x
$$
Your argument is sound as well.
The key is that a solution to $x+\frac{1}{x}=1$ is a 6th root of $1$. Indeed,
$$
x^6-1=(x^3-1)(x^3+1)=(x-1)(x+1)(x^2+x+1)(x^2-x+1)
$$
Thus any root of $x^2-x+1$ is also a root of $x^6-1$.
