Prove that $$\prod_{_{m \mid p_1 \cdots p_n}}\Phi_{2m}(2) = 2^{p_1p_2 \cdots p_n}+1,$$ where $p_1,\ldots,p_n$ are distinct primes.
I was wondering how to use the definitions of cyclotomic polynomials in order to prove this statement. Should we use this definition $$\Phi_n(x) = \prod_{1 \leq k \leq n, \gcd(k,n) = 1}(x-\omega_k^n)?$$
We know that $\prod_{_{m \mid p_1 \cdots p_n}}\Phi_{m}(2) = 2^{p_1p_2 \cdots p_n}-1$, but how do we use this to solve the question?
Suppose we seek to show that
$$\prod_{m|p_1\cdots p_n} \Phi_{2m}(2) = 2^{p_1 p_2\cdots p_n} + 1$$
where the $p_q$ are distinct odd primes. We actually prove the more general result
$$\prod_{m|p_1\cdots p_n} \Phi_{2m}(x) = x^{p_1 p_2\cdots p_n} + 1$$
We start from
$$\Phi_n(x) = \prod_{d|n} (x^d-1)^{\mu(n/d)}.$$
The proof is by induction on $n.$ Note that
$$\Phi_{2p} = (x-1)^{\mu(2p)} (x^2-1)^{\mu(p)} (x^p-1)^{\mu(2)} (x^{2p}-1)^{\mu(1)} \\ = \frac{(x-1)(x^{2p}-1)}{(x^2-1)(x^p-1)}.$$
We get for $n=1$
$$\prod_{m|p_1} \Phi_{2m}(x) = \Phi_{2}(x) \times \Phi_{2{p_1}}(x) = (x+1) \times \frac{(x-1)(x^{2p_1}-1)}{(x^2-1)(x^{p_1}-1)} \\ = x^{p_1} + 1$$
and the base case holds. For the induction we must show that
$$\prod_{m|p_1\cdots p_n} \Phi_{2m}(x) \prod_{m|p_1\cdots p_n} \Phi_{2mp_{n+1}}(x) = x^{p_1 p_2\cdots p_n p_{n+1}} + 1.$$
We have from the definition that with $p$ a prime not dividing $b$ it holds that
$$\Phi_{bp}(x) = \prod_{d|b} (x^d-1)^{\mu(bp/d)} \prod_{d|b} (x^{pd}-1)^{\mu(bp/d/p)} = \frac{\Phi_b(x^p)}{\Phi_b(x)}.$$
This yields for the LHS
$$\prod_{m|p_1\cdots p_n} \Phi_{2m}(x) \prod_{m|p_1\cdots p_n} \frac{\Phi_{2m}(x^{p_{n+1}})}{\Phi_{2m}(x)} \\ = \prod_{m|p_1\cdots p_n} \Phi_{2m}(x^{p_{n+1}}).$$
By the induction hypothesis this is
$$(x^{p_{n+1}})^{p_1 p_2\cdots p_n} + 1$$
and we have the claim.
Remark. Apparently a much simpler proof uses
$$\prod_{d|n} \Phi_d(x) = x^n - 1.$$
We have
$$\prod_{m|p_1\cdots p_n} \Phi_{2m}(x) = \prod_{m|p_1\cdots p_n} \frac{\Phi_m(x^2)}{\Phi_m(x)} = \frac{x^{2p_1p_2\cdots p_n}-1}{x^{p_1p_2\cdots p_n}-1} = x^{p_1p_2\cdots p_n}+1.$$
