If $x+\frac{1}{x}=1$ then $x^2-x+1=0$ and $x\ne -1$ so $(x+1)(x^2-x+1)=x^3+1=0$. Hence $x$ is the cube root of $-1$ so $x^4=-x$ and $x^7=x$. It is now easy to show the values of $x^n+\frac{1}{x^n}$ take the values 1, -1, -2, -1, 1, 2, 1, -1, .... cyclically.

Method 2

${\displaystyle (x+\frac{1}{x})(x^n+\frac{1}{x^n})=x^{n+1}+\frac{1}{x^{n+1}}+x^{n-1}+\frac{1}{x^{n-1}}.}$

But $x+\frac{1}{x}=1$ so

${\displaystyle x^{n+1}+\frac{1}{x^{n+1}}=(x^n+\frac{1}{x^n})-(x^{n-1}+\frac{1}{x^{n-1}}).}$

Now $x^1+\frac{1}{x^1}=1$ and $x^0+\frac{1}{x^0}=2$. Hence

${\displaystyle x^2+\frac{1}{x^2}=1-2=-1}$

${\displaystyle x^3+\frac{1}{x^3}=-1-1=-2}$

${\displaystyle x^4+\frac{1}{x^4}=-2-(-1)=-1}$

${\displaystyle x^5+\frac{1}{x^5}=-1-(-2)=1}$

${\displaystyle x^6+\frac{1}{x^6}=1-(-1)=2}$

${\displaystyle x^7+\frac{1}{x^7}=2-1=1}$

So the process repeats itself in a 6-cycle.

Method 3

Write $x=e^{\log x}=e^y$ and $\mathrm{ny}=\log x^n$. Then $x+\frac{1}{x}=1$ is equivalent to $2\cosh y=1$ and $x^n+\frac{1}{x^n}=2\cosh \mathrm{ny}$. Now ${\sinh }^{2}y=\frac{-3}{4}$ so $\cosh y+\sinh y=e^y=\frac{1-i\sqrt{3}}{2}$. So $e^y$ is at $(1,-\frac{\pi }{3})$ in the Argand diagram.

So all powers of $e^y$ lie on the vertices of a regular hexagon, centre the origin and hence $e^{6y}=1$. So $x^n+\frac{1}{x^n}$ takes values in a 6-cycle.

See Mathematical Gazette December 1969 Number 386