Show that this polynomial (involving n-th roots of unity) has rational coefficients I am tasked with showing that the polynomial $\Phi(x) = \prod_{i=0}^{\phi(n)} (x - \omega_i)$ contains only rational coefficients.  
To give you an example, if $n = 10$, the polynomial is just $\Phi(x) = (x - \omega^1)(x - \omega^3)(x - \omega^7)(x - \omega^9)$.  
Upon working this out, I do get that the coefficients are rational numbers.  
How would I go about proving the general case?  
Some things I know are that the n-th cyclotomic extension (the splitting field for $x^n - 1$) has $\phi(n)$ distinct generators when viewed as a multiplicative group.  I'm not sure how that helps though.    
 A: Suppose we seek to show that
$$\Phi_n(x) = 
\prod_{k=1, \; \gcd(k,n)=1}^n (x - \exp(2\pi i k/n))$$
has integer coefficients.
First observe that for $n\ge 2$
$$\sum_{k=1, \; \gcd(k,n)=1}^n k
= \frac{1}{2} \varphi(n) n.$$
This implies that
$$[x^0] \Phi_n(x) = \Phi_n(0)
= (-1)^{\varphi(n)} \exp(2\pi i \times \varphi(n)n/2/n)
\\ = (-1)^{\varphi(n)} \exp(\pi i \times \varphi(n))
= (-1)^{2\varphi(n)} = 1.$$
Observe that
$$x^n - 1 = \prod_{k=1}^n (x - \exp(2\pi i k/n))
= \prod_{d|n} \prod_{k=1, \; (k,n)=d}^n (x - \exp(2\pi i k/n))
\\ = \prod_{d|n} 
\prod_{q=1, \; (qd,n)=d}^{n/d} (x - \exp(2\pi i qd/n))
= \prod_{d|n} 
\prod_{q=1, \; (q,n/d)=1}^{n/d} (x - \exp(2\pi i q/(n/d)))
\\ = \prod_{d|n} \Phi_{n/d}(x) = \prod_{d|n} \Phi_d(x).$$ 
We now prove the claim by induction, it holds when $n=1$ where we have
$\Phi_1(x) = x - 1.$ Write
$$x^n - 1 = \Phi_n(x) \prod_{d|n, \; d \lt n} \Phi_d(x)
= \Phi_n(x) g_n(x).$$
Note that the constant coefficient of $g_n(x)$ is
$$[x^0] g_n(x) = \prod_{d|n, \; d \lt n} \Phi_d(0)
= -1 \times \prod_{d|n, \; 1\lt d \lt n} \Phi_d(0)
= -1.$$
Let $\Phi_n(x)  =  \sum_{q=0}^{\varphi(n)} c_q x^q$ 
and $g_n(x) = \sum_{q=0}^{n-\varphi(n)} a_q x^q$
and suppose  to the contrary  that there is a  non-integer coefficient
$c_k = [x^k] \Phi_n(x)$ in  $\Phi_n(x)$ with $k$ being the smallest to
produce such a  coefficient. Here $k\ge 1$ since  we know the constant
coefficient is an integer.
This would imply
$$[x^k] (x^n-1) = [x^k] \Phi_n(x) g_n(x)
= \sum_{q=0}^{\min(k,\varphi(n))} c_q a_{k-q}.$$
With $k\le \varphi(n)$ by definition we thus have
$$[x^k] (x^n-1)
= \sum_{q=0}^{k} c_q a_{k-q}.$$
The  first $k$ terms  in this  sum are  integers by  construction (the
$c_q$ precede $c_k$ and the  $a_{k-q}$ are integers by induction). The
last term is $c_k [x^0] g_n(x) =  - c_k,$ which is not an integer. The
sum of a  series of integers and a single  non-integer fraction is not
an integer. We now have a contradiction because the coefficient $[x^k]
(x^n-1)$ is always an integer, QED.  (Here some of the $a_{k-q}$ could
potentially  be  zero for  small  $q$ but  this  does  not affect  the
argument.)
Source. This  introductory lecture  / textbook style  proof was
found at the following 
Internet link.
