Show that if $\operatorname{char}(k)=0$, then $f_d(x)=\prod_{c|d}(x^{d/c}-1)^{\mu(c)}$. I'm trying to show that if $k$ is a field of characteristic zero, then 
$$f_d(x)=\prod_{c|d}(x^{d/c}-1)^{\mu(c)}$$
where $f_d$ is the cyclotomic polynomial of degree $d$, $f_d(x)=\prod(x-\lambda)$ where $\lambda$ runs over primitive $d$th roots of unity, and $\mu$ is the Mobius function, i.e. $\mu(c)$ is the sum of the primitive $c$th roots of unity.
I don't really know what the best way to approach this problem is. My first thought was to just show that all primitive $d$th roots of unity are roots of the right hand side, then compare orders of the polynomials. We can choose one primitive $d$th root of unity, call it $\zeta$, so that all primitive $d$th roots are given by $\zeta^k$ where $(k,d)=1$ and $0\le k<d$; then I need to show that there exists some $c$ dividing $d$ with $(x^{d/c}-1)^{\mu(c)}=0$. But this isn't exactly working very well.
Does anybody have a suggestion on a better way I could be doing this?
 A: If $k$ is a field of characteristic $0$, then it contains $\mathbb{Q}$, and the cyclotomic polynomials in $k[x]$ are the same as the usual ones.
Now in $\overline{k}$ (the algebraic closure) for every $n$ there is a primitive $n$th root of unity, and you can define $\Phi_n$ as
$$\Phi_n(x) = \prod_{m=1,gcd(n,m)=1}^n (x-\zeta_n^m)$$
It is not obvious that this is a polynomial with rational coefficients, until you write that
$$x^n-1 = \prod_{k=1}^n(x-\zeta_n^k) = \prod_{d=1}^n\prod_{k=1,gcd(n,k)=d}^n(x-\zeta_n^k)=\prod_{d| n}\prod_{m=1,gcd(n/d,m)=1}^{n/d}(x-\zeta_{dn/d}^{dm})$$
$$ =\prod_{d| n}\prod_{m=1,gcd(n/d,m)=1}^{n/d}(x-\zeta_{n/d}^{m})= \prod_{d | n}\Phi_{n/d}(x)$$
So that with $n$ the least index such that $ \Phi_n(x) \not \in \mathbb{Q}[x]$, 
you get that $x^n-1 = \prod_{d | n}\Phi_{n/d}(x) =\Phi_n(x) P(x)$ where $P(x) \in \mathbb{Q}[x] $ and $\Phi_n(x) \not \in \mathbb{Q}[x]$,
i.e. $x^n -1 \not\in \mathbb{Q}[x]$, a contradiction.
Finally, with the definition of the Möbius function $\sum_{d | n}\mu(d)=1_{n = 1}$ you get 
$$\prod_{c | n}(x^{n/c}-1)^{\mu(c)} = \prod_{c | n} \prod_{d | c}\Phi_{c/d}(x)^{\mu(c)}= \prod_{e | n}\Phi_{e}(x)^{\sum_{c | n/e} \mu(c)} = \Phi_n(x)$$
It remains to show that an other possible definition is $\mu(n) = \sum_{m=1,gcd(n,m)=1}^n \zeta_n^m$, that you get from
$$\sum_{d | n} \sum_{m=1,gcd(d,m)=1}^n \zeta_{d}^m =\sum_{d | n} \sum_{m=1,gcd(n/d,m)=1}^n \zeta_{n/d}^m= \sum_{d | n} \sum_{m=1,gcd(n/d,m)=1}^n \zeta_{n}^{md}$$ $$= \sum_{k=1}^n \zeta_n^k = 1_{n=1}$$
proving that $\displaystyle\sum_{m=1,gcd(n,m)=1}^n \zeta_n^m = \mu(n)$
