# Degree of a formal power series involving Mobius function

I am reading Enumerative Combinatorics by Richard Stanley, and I came across the following expression: $$(1-x^n)^{\frac{-\mu(n)}{n}}$$, where $$\mu(n)$$ is the usual Mobius function from number theory.

This expression can also be written as:

$$(1-x^n)^{\frac{-\mu(n)}{n}}=\sum_{i \ge 0} \binom{-\mu(n)/n}{i}(-1)^ix^{in}$$

This part is clear. But then the author claims that the expansion above shows that $$(1-x^n)^{\frac{-\mu(n)}{n}}=1+H(x)$$, where $$H(x)$$ is a formal power series of degree $$n$$. Why is that?

(Also: the degree of a formal power series is defined here as the smallest integer for which the corresponding coefficient is nonzero.

$$\binom{-\mu(n)/n}{i}$$ is defined in the usual way, as $$\frac{(-\mu(n)-i+1)\dots (-\mu(n))}{i!}$$)

As $$\mu(n)\in\{0,1,-1\}$$ we have one of three cases.
If $$\mu(n)=0$$ then $$(1-x^n)^{\mu(n)/n}=1$$ so that $$H(x)=0$$.
If $$\mu(n)=1$$ then $$(1-x^n)^{\mu(n)/n}=(1-x^n)^{1/n}=1-\frac{x^n}{n}-(n-1)\frac{x^{2n}}{2n^2} -\cdots$$ so that $$H(x)=-\frac{x^n}{n}+\text{higher terms}.$$
If $$\mu(n)=-1$$ then $$(1-x^n)^{\mu(n)/n}=(1-x^n)^{-1/n}=1+\frac{x^n}{n}+(n+1)\frac{x^{2n}}{2n^2} -\cdots$$ so that $$H(x)=\frac{x^n}{n}+\text{higher terms}.$$