Show there is a tower of cyclic field extensions of prime degree from $\mathbb{Q}(\sqrt[12]{5})$ to $\mathbb{Q}$

Question

Of course, it suffices to find tower of Galois extensions of prime degree, as these would have to be cyclic. My first thought was to try extending $\mathbb{Q}$ first by $\sqrt 5$, then $\mathbb{Q}(\sqrt 5)$ by $\sqrt[4]{5}$ and $\mathbb{Q}(\sqrt[4]{5})$ by $\sqrt[12]{5}$, but then the last extension isn't Galois as it's not normal. I tried extending $\mathbb{Q}(\sqrt[4]{5})$ by $\omega_3\sqrt[12]{5}$ where $\omega_3$ denotes the third root of the unit, but of course this gives me a different field at the top of the tower, not $\mathbb{Q}(\sqrt[12]{5})$. In general, I'm clueless on how to find a Galois extension of degree three which doesn't have imaginary numbers.