What is the discriminant of $\Phi_{2^n}(X)$ over $\Bbb Q$ for $n=1$ since $\Phi_2(X)=X+1$ has only one root $-1$?
I have calculated $disc(\Phi_{2^n}(X))$ for $n\geq 2$ which matches exactly with what happens for odd primes, i.e., $disc(\Phi_{p^n}(X))=(-1)^{\frac{d(d-1)}{2}}p^{p^{n-1}(np-n-1)}$ with $d=\phi(p^n)=p^{n-1}(p-1)$. Only difference is for odd prime $p$, $N_{\Bbb Q(\zeta_p)/\Bbb Q}(\zeta_p-1)=p$ but for $p=2$, $N_{\Bbb Q(\zeta_2)/\Bbb Q}(\zeta_2-1)=-2$, noting that $\Bbb Q_{\zeta_2}=\Bbb Q$ and $\zeta_2=-1$. But latter on even power takes care of the calculation.
Yet I don't understand what should be the convention to take the discriminant of $\Phi_2(X)$. My guess is $-1$, since for odd prime $p$, $disc(\Phi_{p}(X))=(-1)^{p \choose 2}p^{p-2}$. If we take $p=2$ in this expression we get $-1$. Any information on this will be helpful for me. Thanks.