Cyclotomic polynomials satisfy $x^n-1=\prod_{n_1|n}\phi_{n_1}(X)$
Prove that $\phi_{2n}(X)=\phi_n(-X)$
I need to be able to show that: $\zeta$ is an $n$-th primitive root of unity then $-\zeta$ is a $2n$-th root of unity
Let: $2n=p_1p_2..p_k$ and $n=p_1p_2..p_k$
Using the above formula for each case:
$$x^{2n}-1=\phi_{2}(X)\phi_{p_1}(X)...\phi_{p_k}(X)\phi_{2n}(X)$$
$$x^{n}-1=\phi_{p_1}(X)...\phi_{p_k}(X)\phi_{n}(X)$$
So $\phi_{2n}(X)=\frac{x^{2n}-1}{\phi_{2}(X)\phi_{p_1}(X)...\phi_{p_k}(X)}$ and $\phi_{n}(X)=\frac{x^{n}-1}{\phi_{p_1}(X)...\phi_{p_k}(X)}$
So $\phi_{n}(-X)=\frac{-x^{n}-1}{\phi_{p_1}(-X)...\phi_{p_k}(-X)}$
I cannot see how $\phi_{2n}(X)=\phi_n(-X)$ from here. I tried to think instead in terms of $\zeta$ and $-\zeta$ but I am unsure
Could someone help me prove this?
Also, say we want to find the cyclotomic polynomial when $n$ is large and even, is using the above identity the best way to calculate it? For example, say we wanted to find $\phi_{50}(X)$, would we first calculate $\phi_{25}(X)$, or is there a quicker way?