Let $p$ be a prime and $\zeta_{p^r}$ be a primitive $p^r$-th root of unity. Then many textbooks, e.g. Milner's, ask one to prove that $\mbox{Gal}(\mathbb{Q}(\zeta_{p^\infty})/\mathbb{Q})\cong \mathbb{Z}_p$ where $\mathbb{Q}(\zeta_{p^\infty})=\cup_{i\geq 1} \mathbb{Q}(\zeta_{p^i})$.
But as I try to understand it, I know that $\mbox{Gal}(\mathbb{Q}(\zeta_{p^n})/\mathbb{Q})\cong (\mathbb{Z}/p^n\mathbb{Z})^\times$.
And so when $p=2$, we have $\mbox{Gal}(\mathbb{Q}(\zeta_{2^n})/\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2^{n-2}\mathbb{Z}$ and so $$\mbox{Gal}(\mathbb{Q}(\zeta_{2^\infty})/\mathbb{Q})\cong \varprojlim_{n\geq 1}(\mathbb{Z}/2^n\mathbb{Z})^\times\cong \varprojlim_{n\geq 1} (\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2^{n-2}\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}_2 (\cong \mathbb{Z}_2^\times).$$
When $p\geq 3$, we have $\mbox{Gal}(\mathbb{Q}(\zeta_{p^n})/\mathbb{Q})\cong \mathbb{Z}/(p-1)\mathbb{Z}\times \mathbb{Z}/p^{n-1}\mathbb{Z}$ since $(\mathbb{Z}/p^n\mathbb{Z})^\times$ is cyclic of order $p^{n-1}(p-1)$ and so \begin{align*} \mbox{Gal}(\mathbb{Q}(\zeta_{p^\infty})/\mathbb{Q})\cong & \varprojlim_{n\geq 1}(\mathbb{Z}/p^n\mathbb{Z})^\times \\ \cong & \varprojlim_{n\geq 1} (\mathbb{Z}/(p-1)\mathbb{Z}\times \mathbb{Z}/p^{n-1}\mathbb{Z})\\ \cong & \mathbb{Z}/(p-1)\mathbb{Z}\times \mathbb{Z}_p(\cong \mathbb{Z}_p^\times). \end{align*}
Surely, I (mis)use that $\varprojlim$ is exchangeable with group product (and taking units). But surely, $\mathbb{Z}_{p}$ and $\mathbb{Z}_p^\times$ are not the same.
Can anyone explain where I go wrong?
You are correct. The correct statement should be $$\operatorname{Gal}(\mathbf{Q}(\zeta_{p^\infty})/\mathbf{Q})\cong \mathbf{Z}_p^\times.$$
for $p$ prime. The (projective) limit commutes with taking unit groups. This is a special case of the fact that the functor $$\mathbf{Ring}\to \mathbf{Grp},R\mapsto R^\times$$ is a right adjoint and that right adjoints commute with taking limits.