cyclotomic polynomial $\Phi_{2n}(x)$

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Question is:

"Show that $\Phi_{2n}(x)=\Phi_n(-x)$ for all odd numbers $n>1$"

I try to prove this as follow,

$\prod_{d \mid 2n} \Phi_d(x) =x^{2n}-1 = (x^n-1)(x^n+1) = -(x^n-1)((-x)^n-1) = -\prod_{d \mid n} \Phi_d(x)\Phi_d(-x)$

Then I write,

$ \Phi_n(x)\Phi_{2n}(x)\prod_{d \mid n; d\neq n}\Phi_d(x)\Phi_{2d}(x) = - \Phi_n(x)\Phi_{n}(-x)\prod_{d \mid n;d\neq n} \Phi_d(x)\Phi_d(-x)$

Then

$ \Phi_{2n}(x)\prod_{d \mid n; d\neq n}\Phi_{2d}(x) = - \Phi_{n}(-x)\prod_{d \mid n; d\neq n} \Phi_d(-x)$

My claim is $ \prod_{d \mid n; d\neq n}\Phi_{2d}(x) = - \prod_{d \mid n; d\neq n} \Phi_d(-x)$

But I did not prove this claim.. Any idea?

or dou you have another proof?