I recently found a very short proof of the prime power case in the incredibly enlightening notes of Peter Stevenhagen: Number rings.
The proof follows almost entirely from the Dedekind–Kummer theorem for orders $\mathbb{Z}[\alpha]$ (Theorem 3.1 in the notes). The theorem states that if $f(x) \in \mathbb{Z}[x]$ is the minimal polynomial of $\alpha$ and $\overline{f} = \prod \overline{g}_i^{e_i}$ mod $p$, then the primes $\mathfrak{p}$ above $p$ in $\mathbb{Z}[\alpha]$ correspond exactly to the $\overline{g_i}$. Moreover, $\mathfrak{p}$ is 'singular' iff $e_i > 1$ and also $p^2$ divides the remainder $r$ in $f = ag_i + r$. I have left out some details which are in Stevenhagen's notes.
To apply this to $\mathbb{Z}[\zeta_{p^k}]$, write $\Phi_{p^k}(x) = \frac{x^{p^k} - 1}{x^{p^{k-1}}-1}$. Mod $p$, this is just $\frac{(x-1)^{p^k}}{(x-1)^{p^{k-1}}} = (x-1)^{p^k - p^{k-1}}$ which shows $\mathfrak{p} = (\zeta - 1, p)$ is the only prime above $p$ and also since the remainder of $\Phi_{p^k}$ divided by $x-1$ is just $\Phi_{p^k}(1) = p$ (which is not divisible by $p^2$), it follows that $\mathfrak{p}$ is 'regular'. It is easy to check $\Phi_{p^k}$ and its derivative are coprime mod $q$ for all other primes $q$, implying all the $e_i$'s are 1. Thus, all the primes above $q$ are 'regular' as well (and unramified). All primes being regular implies the ring is Dedekind, so $\mathbb{Z}[\zeta]$ must be the ring of integers. This is covered in more detail in Theorem 3.12 in the notes.
This argument still works prime-by-prime but avoids having to localize (of course, the localization is hidden under the hood in the proof of Dedekind–Kummer). Moreover, this approach is just a specific application of the very general framework of applying Dedekind–Kummer to try to determine the ring of integers containing some $\mathbb{Z}[\alpha]$ and the same exact type of analysis will quickly determine the ring of integers of $\mathbb{Q}[\sqrt{d}]$ and $\mathbb{Q}[\sqrt[3]{d}]$, just to name a few.
