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Ira Gessel
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It's very easy (and this is probably why people are telling you that the question is not appropriate for MO).

You have a certain integer $X$, and you'd like to prove that $X$ is divisible by $\Phi_n$.

You know that $v_p(X)\geq\rho_0(n/p)$ for every odd prime $2\sqrt n\leq p\leq n$ (from the "Notations" section), meaning that $X$ is divisible by the $\rho_0(n/p)$-th power of $p$.

Hence youryou're done.

It's very easy (and this is probably why people are telling you that the question is not appropriate for MO).

You have a certain integer $X$, and you'd like to prove that $X$ is divisible by $\Phi_n$.

You know that $v_p(X)\geq\rho_0(n/p)$ for every odd prime $2\sqrt n\leq p\leq n$ (from the "Notations" section), meaning that $X$ is divisible by the $\rho_0(n/p)$-th power of $p$.

Hence your done.

It's very easy (and this is probably why people are telling you that the question is not appropriate for MO).

You have a certain integer $X$, and you'd like to prove that $X$ is divisible by $\Phi_n$.

You know that $v_p(X)\geq\rho_0(n/p)$ for every odd prime $2\sqrt n\leq p\leq n$ (from the "Notations" section), meaning that $X$ is divisible by the $\rho_0(n/p)$-th power of $p$.

Hence you're done.

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DamienC
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It's very easy (and this is probably why people are telling you that the question is not appropriate for MO).

You have a certain integer $X$, and you'd like to prove that $X$ is divisible by $\Phi_n$.

You know that $v_p(X)\geq\rho_0(n/p)$ for every odd prime $2\sqrt n\leq p\leq n$ (from the "Notations" section), meaning that $X$ is divisible by the $\rho_0(n/p)$-th power of $p$.

Hence your done.