This is a statement made by my professor, but I suspect it to be false.
He said that if $k(\alpha)$ contains a primitive nth root of unity and $\alpha^n = a \in k$ then $k(\alpha)$ is a cyclic extension
I don't think this is true, consider the case $\alpha$ as a primitive $nth$ root, and $k = \mathbb{Q}$, then we know that this field extension is not always cyclic. right?
The reason he wanted this statement is because we are trying to prove that $k$ adjoinin a bunch of $ith$ roots of unity where $i$ is in a finite subset of $\mathbb{N}$ is a root extension with each successive intermediate field extension $K_{i+1}/K_i$ is cyclic.
Some counterexamples to the statement you posed:
$$\mathrm{Gal}(\mathbb{Q}(\zeta_8) / \mathbb{Q}) \cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} $$ $$\mathrm{Gal}(\mathbb{Q}(\zeta_{12}) / \mathbb{Q}) \cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z} $$ $$\mathrm{Gal}(\mathbb{Q}(\zeta_{15}) / \mathbb{Q}) \cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 4 \mathbb{Z} $$
More generally, it turns out that there is an isomorphism
$$ \mathrm{Gal}(\mathbb{Q}(\zeta_n) / \mathbb{Q}) \cong (\mathbb{Z} / n \mathbb{Z})^\times $$
where by $R^\times$ I mean the unit group of the ring $R$.
However, adjoining roots of unity always gives abelian extensions, and so they can always be decomposed into a tower of cyclic extensions; e.g. these examples can be decomposed into
$$\mathbb{Q}(\zeta_8) / \mathbb{Q}(\zeta_4) / \mathbb{Q} $$ $$\mathbb{Q}(\zeta_{12}) / \mathbb{Q}(\zeta_4) / \mathbb{Q} $$ $$\mathbb{Q}(\zeta_{15}) / \mathbb{Q}(\zeta_3) / \mathbb{Q} $$