<s>[INST] Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that 
$x^5 + \frac{1}{x^5} = 1$ as well.
 [/INST] Notice that $x+x^{-1}=1$ implies that
$$1 = (x+x^{-1})^3 = (x^3+x^{-3}) + 3(x+x^{-1}),$$
and thus
$$x^3+x^{-3} = -2.$$
Therefore,
$$1 = (x+x^{-1})^5 = (x^5+x^{-5}) + 5(x^3+x^{-3}) + 10(x+x^{-1}),$$
giving
$$x^5+x^{-5} = 1.$$
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