Is there a better proof of this fact in number theory/formal group theory? Let $\Phi_n$ be the $n$'th cyclotomic polynomial, and put 
\begin{align*}
 a_n &= \Phi_n(1) \\
 b_n &= \gcd\left(\left(\begin{array}{c} n \\ 1\end{array}\right),\dotsc,\left(\begin{array}{c} n \\ n-1\end{array}\right)\right) \\
 c_n &= \begin{cases}
         p & \text{ if } n = p^k \text{ for some prime } p \text{ and } k>0 \\
         1 & \text{ otherwise. } 
        \end{cases}
\end{align*}
It is well-known that $a_n=b_n=c_n$.  Indeed, there are a bunch of ways to prove that $a_n=c_n$, and a bunch of ways to prove that $b_n=c_n$.  I ask: is there a more direct proof that $a_n=b_n$?  A good answer might give a tidier approach to some fundamental results in formal group theory.  Ideally I'd like a proof that expresses $a_n$ as a $\mathbb{Z}$-linear combination of binomial coefficients as in $b_n$.
 A: Inspired by the previous answer I am thinking about simplest possible ways to prove
$$
gcd\left(\binom n1_q,...,\binom n{n-1}_q\right)=\Phi_n(q).
$$
One most primitive way to do it is to check that both sides have the same roots. On the left, multiplicity of a primitive root of unity of degree $k$ is
$$
\min\left\{\left[\frac nk\right]-\left[\frac mk\right]-\left[\frac{n-m}k\right]\ |\ m=1,...,n-1\right\}
$$
and it must not be difficult to show that this is 0 for $k\ne n$ and 1 for $k=n$.
A: This answer seemed to be a simplification of arguments given by Aaron Meyerowitz. 
As it was mentioned numbers $a_n=\Phi_n(1)$ are uniquely determined by $$n=\prod_{d \mid n}a_d.$$ So it is sufficient to check that numbers $c_n$ satisfy the same equation. But $c_n=e^{\Lambda(n)}$, where
$$\Lambda(n)=\begin{cases}
\log p & \text{ if }n = p^k, \\
0 & \text{otherwise},
\end{cases}$$
and verification of identity $n=\prod_{d \mid n}c_d$
is an easy exercise.
We can express $a_n$ as a $\mathbb{Z}$-linear combination of binomial coefficients as in $b_n$ in the following way (this construction is taken from the paper Coefficient rings of formal groups).
If $n=p^k$ then $\binom{n}{p^{k-1}}\equiv p\pmod{p^2},$ so we can easely find $\lambda_{p^{k-1}}$ such that $\lambda_{p^{k-1}}\binom{p^k}{p^{k-1}}\equiv p\pmod {p^{k}}$. So for some $\lambda_{1}$
$$\lambda_{p^{k-1}}\binom{p^k}{p^{k-1}}+\lambda_{1}\binom{p^k}{1}=p.$$
Now let $n=p_1^{k_1}\ldots p_s^{k_s}$, where $s>1$. Then by Kummer's theorem $\mathrm{ord}_{p_i}\binom{n}{p_i^{k_i}}=0$ and $\mathrm{ord}_{p_j}\binom{n}{p_i^{k_i}}\ge k_j$ ($j\ne i$). Taking $\lambda_{p_i^{k_i}}\equiv \binom{n}{p_i^{k_i}}^{-1}\pmod{p_i^{k_i}}$ we'll have
$$\lambda_{p_1^{k_1}}\binom{n}{p_1^{k_1}}+\ldots+\lambda_{p_s^{k_s}}\binom{n}{p_s^{k_s}}\equiv 1\pmod n.$$
So for some $\lambda_1$ 
$$\lambda_{p_1^{k_1}}\binom{n}{p_1^{k_1}}+\ldots+\lambda_{p_s^{k_s}}\binom{n}{p_s^{k_s}}+\lambda_{1}\binom{n}{1}=1.$$
