Proving the Möbius formula for cyclotomic polynomials We want to prove that 
$$ \Phi_n(x) = \prod_{d|n} \left( x^{\frac{n}{d}} - 1 \right)^{\mu(d)} $$
where $\Phi_n(x)$ in the n-th cyclotomic polynomial and $\mu(d)$ is the Möbius function defined on the natural numbers.
We were instructed to do it by the following stages:
Using induction we assume that the formula is true for $n$ and we want to prove it for $m = n p^k$ where $p$ is a prime number such that $p\not{|}n$.
a) Prove that $$\prod_{\xi \in C_{p^k}}\xi = (-1)^{\phi(p^k)} $$ where $C_{p^k}$ is the set of all primitive $p^k$-th roots of unity, and $\phi$ is the Euler function. I proved that.
b) Using the induction hypothesis show that 
$$ \Phi_m(x) = (-1)^{\phi(p^k)}  \prod_{d|n} \left[ \prod_{\xi \in C_{p^k}} \left( (\xi^{-1}x)^{\frac{n}{d}} - 1 \right) \right]^{\mu(d)} $$
c) Show that
$$ \prod_{\xi \in C_{p^k}} \left( (\xi^{-1}x)^{\frac{n}{d}} - 1 \right) = (-1)^{\phi(p^k)}   \frac{x^{\frac{m}{d}}-1}{x^{\frac{m}{pd}} - 1} $$
d) Use these results to prove the formula by substituting c) into b).
I am stuck in b) and c).
In b) I tried to use the recursion formula $$ x^m - 1 = \prod_{d|m}\Phi_d(x) $$ and 
$$ \Phi_m(x) = \frac{x^m-1}{ \prod_{\stackrel{d|m}{d<m}} \Phi_d(x)} . $$
In c) I tried expanding the product by Newton's binom using $\phi(p^k) = p^k ( 1 - 1/p)$. I also tried replacing the product by $\xi \mapsto [ \exp(i2\pi / p^k) ]^j$ and let $j$ run on numbers that don't divide $p^k$. In both way I got stuck.
I would appreciate help here.

I found a solution that is easy to understand for those who want to know how to solve without following the steps given in the problem. 
Click to see the source. 
First, we have a formula
$$
{x^{n}-1=\Pi_{d|n}\Phi_{d}(x)}
$$
Then, by taking the logarithm on the both sides, 
$$
\log(x^{n}-1)=\log(\Pi_{d|n}\Phi_{d}(x))=\Sigma_{d|n}(\log\Phi_{d}(x))
$$
Now, we use the Mobius Inversion Formula by taking $f_{n}=\log\Phi_{n}$ and $F_{n}=\Sigma_{d|n}(\log\Phi_{d})$.
\begin{align*}
\log(\Phi_{n}(x)) & = \Sigma_{d|n}\mu(\frac{n}{d})\log(x^{d}-1)\\
                  & = \Sigma_{d|n}\log(x^{d}-1)^{\mu(\frac{n}{d})}\\
                  & = \log\Pi_{d|n}(x^{d}-1)^{\mu(\frac{n}{d})}
\end{align*}
Hence, we have 
$$
\Phi_{n}(x)=\Pi_{d|n}(x^{d}-1)^{\mu(\frac{n}{d})}
$$
