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 submitted paper [Coefficient rings of formal groups][1]).

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][2] $\mathrm{ord}_{p_i}\binom{n}{p_i^{k_i}}=0$ 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 $\lambda_1$ 
$$\lambda_{p_1^{k_1}}\binom{n}{p_1^{k_1}}+\ldots+\lambda_{p_s^{k_s}}\binom{n}{p_s^{k_s}}+\lambda_{1}\binom{n}{1}=1.$$


  [1]: http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=8523&option_lang=eng
  [2]: https://en.wikipedia.org/wiki/Kummer%27s_theorem