$\Phi_{2^n}(x) = x^{2^{n-1}}+1$ by Mobius Inversion

Question

I want to prove that $\Phi_{2^n}(x) = x^{2^{n-1}}+1$.

This is not hard to do by using recursion. However, I want to know if there's a simpler way to do this using the Mobius Inversion formula, which states as follows:

If $a_n = \prod_{d|n} b_d$, then $b_n = \prod_{d|n}(a_{n/d})^{\mu(d)}$, where

$$\mu(d) = \left\{ \begin{array}{lcc} 1, & d=1 \\ \\ 0, &\mbox{ d is not square free} \\ \\ (-1)^l, &\mbox{$n=p_1p_2\dots p_l$, where $p_i$ distinct} \end{array} \right.$$

Answer 1

Recall that $$(x^n-1)=\prod_{d\mid n}\Phi_d(x).$$ By Möbius Inversion, this implies $$\Phi_n(x)=\prod_{d\mid n}\left(x^d-1\right)^{\mu\left({n\over d}\right)}.$$ In particular, $$\Phi_{2^n}(x)=\prod_{d\mid 2^n}\left(x^d-1\right)^{\mu\left({2^n\over d}\right)}=\frac{x^{2^n}-1}{x^{2^{n-1}}-1}=x^{2^{n-1}}+1,$$ just as we wanted to prove. $\blacksquare$