2
Argument
Let, $f(x) = \sum_{r=1}^\infty \mu(r) x^r$ where $\mu(r)$ is the mobius function. Hence,
$$ f(x) + f(x^2) + \dots = x$$
Now, let $x \to x^2$:
$$ 0+f(x^2) + 0+\dots = x^2$$
Similarly $x \to x^3$ :
$$ 0 + 0 + f(x^3) + \dots = x^3 $$
$$\vdots$$
$$ \underbrace{0 + 0}_{n-1 \text{ ti...
