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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).

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 $\lambda_1$ $$\lambda_{p_1^{k_1}}\binom{n}{p_1^{k_1}}+\ldots+\lambda_{p_s^{k_s}}\binom{n}{p_s^{k_s}}+\lambda_{1}\binom{n}{1}=1.$$

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 $\lambda_1$ $$\lambda_{p_1^{k_1}}\binom{n}{p_1^{k_1}}+\ldots+\lambda_{p_s^{k_s}}\binom{n}{p_s^{k_s}}+\lambda_{1}\binom{n}{1}=1.$$

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).

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 $n=p_1^{k_1}\ldots p_s^{k_s}$, где $s>1$. 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 $\lambda_1$ $$\lambda_{p_1^{k_1}}\binom{n}{p_1^{k_1}}+\ldots+\lambda_{p_s^{k_s}}\binom{n}{p_s^{k_s}}+\lambda_{1}\binom{n}{1}=1.$$

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).

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 $\lambda_1$ $$\lambda_{p_1^{k_1}}\binom{n}{p_1^{k_1}}+\ldots+\lambda_{p_s^{k_s}}\binom{n}{p_s^{k_s}}+\lambda_{1}\binom{n}{1}=1.$$

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.

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).

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 $n=p_1^{k_1}\ldots p_s^{k_s}$, где $s>1$. 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 $\lambda_1$ $$\lambda_{p_1^{k_1}}\binom{n}{p_1^{k_1}}+\ldots+\lambda_{p_s^{k_s}}\binom{n}{p_s^{k_s}}+\lambda_{1}\binom{n}{1}=1.$$