For $n\geq 2$, show that $$X^{\varphi(n)}\Phi_{n}(X^{-1})=\Phi_{n}(X)$$ where $\varphi$ is Euler totient function, $\Phi_{n}$ is the $n$th cyclotomic polynomial.
I've tried some discrete examples for some $n$, it all matches our result. And we have several formulas for the cyclotomic polynomial. But I don't know which one is most useful in our case?
[edited to correct statement]
$\Phi_n(X)$ is the product of all $(X-\zeta)$ where $\zeta$ is an $n$-th root of unity but not a $d$-th root of unity for any proper divisor $d$ of $n$. Hence the degree of the $n$-th cyclotomic polynomial is $\varphi(n)$, the number of $k \in \left\{1,2,\ldots,n\right\}$ that are coprime with $n$.
Now $\zeta^k=1$ if and only if $(\zeta^{-1})^k=1$, so that $\zeta^{-1}$ is a root of $\Phi_n(X)$ whenever $\zeta$ is a root.
Now both $X^{\varphi(n)}\Phi_n(X^{-1})$ and $\Phi_n(X)$ are monic polynomials of the same degree with distinct roots, and every root $\zeta^{-1}$ of the first is a root of the second. Hence they are equal.