Context: I am currently reading through Serge Lang's Algebraic Number Theory without much knowledge of category theory or advanced algebra.

In the book, $\mathbb{Z}_p$ is defined as a subgroup of infinite direkt product $$\mathbb{Z}_p = \cdots \times \mathbb{Z}/p^n\mathbb{Z}\times\mathbb{Z}/p^{n-1}\mathbb{Z}\times\cdots\times \mathbb{Z}/p\mathbb{Z}$$ where the $n$-th "component of the sequence" modulo $p^{n-1}$ equals the $n-1$-th component, i.e. $x_n \equiv x_{n-1}~~(\text{mod}~p^{n-1})$. This process of successively taking products is formalized as the (projective) limit $$\mathbb{Z}_p = \lim_\leftarrow ~\mathbb{Z}/p^n\mathbb{Z}~.$$

Question: Later on the groups $U = \mathbb{Z}_p^\times$ and $U_n = 1+p^n\mathbb{Z}_p$ are introduced and it is shown that $U/U_n \cong (\mathbb{Z}/p^n\mathbb{Z})^\times$.

Then the book states (without further explanation) that $$U = \lim_\leftarrow~U/U_n~.$$ If have trouble understanding why this is true. It feels like it should be true but from a formal point of view I need that $$\lim_\leftarrow~(\mathbb{Z}/p^n\mathbb{Z})^\times~=~\big(\lim_\leftarrow~\mathbb{Z}/p^n\mathbb{Z}\big)^\times$$ to be able to resort to the definition of $\mathbb{Z}_p$. Is there an elementary way to explain why this or the above statement about $U$ is true without using much theory about projective limits, profinite groups and related concepts?


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