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.