I'll follow up my comment with a sketch of how I assume the exercise is meant to be solved. People with knowledge in $p$-adics will see I try to recover the standard (for $p \neq 2$) $\Bbb Z_p^\times \simeq \mu(\Bbb Q_p) \times (1+p\Bbb Z_p)$ -- and Teichmüller representatives/roots of unity on the first factor, and the second factor being $\simeq (1+p)^{\Bbb Z_p} \simeq (\Bbb Z_p, +)$ -- with arithmetic congruences modulo $p$-powers.
Let's look at the rings $\Bbb Z/p^j\Bbb Z$. Everyone since Euler knows that the unit group $(\Bbb Z/p^j\Bbb Z)^\times$ has order $(p-1)\cdot p^{j-1}$, which suggests it's of the form $C_{p-1} \times C_{p^{j-1}}$, where $C_n$ denotes the cyclic group of order $n$. (That's actually not the case for $p=2$ ($j\ge 3$), see below, but indeed for all other primes.) Well great: If that is the case, and furthermore if we can establish these isomorphisms in a compatible way when we let $j$ vary, then we have
$$\Bbb Z_p^\times \simeq\varprojlim (\Bbb Z/p^j\Bbb Z)^\times \simeq \varprojlim (C_{p-1} \times C_{p^j}) \simeq \varprojlim C_{p-1} \times \varprojlim C_{p^j} \simeq C_{p-1} \times \Bbb Z_p$$
(there's a few things to check here maybe, but this chain of isomorphisms should be straightforward).
For $p \neq 2$, such compatible isomorphisms $(\Bbb Z/p^j\Bbb Z)^\times \simeq C_{p-1} \times C_{p^{j-1}}$can be found like this:
- First, the $C_{p^{j-1}}$-part. I claim that the element $1+p$ (formally, its various residues mod $p^j$) generates such a cyclic subgroup. Namely, Let $y_j=$ the residue of $1+p$ in $\Bbb Z/p^j\Bbb Z$.
Check that the order of $y_j$ in $(\Bbb Z/p^j\Bbb Z)^\times$ is $p^{j-1}$, i.e. that $(1+p)^{p^k} \not \equiv 1$ mod $p^{j}$ for $k <j-1$, but $(1+p)^{p^{j-1}} \equiv 1$ mod $p^{j}$. For this, you need some playing with $p$-divisibility of binomial coefficients $\binom{p^k}{n}$, and at some point you really need $p \neq 2$.
Compatibility is clear, as the projection $(\Bbb Z/p^j\Bbb Z) \rightarrow (\Bbb Z/p^i\Bbb Z)$ sends $y_j$ to $y_i$.
- The $C_{p-1}$ part needs different considerations. A crucial fact here, proved by induction, is
$$a \equiv b \text{ mod } p \Rightarrow a^{p^{j-1}} \equiv b^{p^{j-1}} \text{ mod } p^j \quad (*)$$
With this, for $a \in\lbrace 1, ..., p-1\rbrace$, check that the residues of $a^{p^{j-1}}$ in $\Bbb Z/p^j\Bbb Z$ are distinct from each other and form a subgroup isomorphic to $(\Bbb Z/p\Bbb Z)^\times$, which is cyclic of order $p-1$.
Choose and fix $2 \le a \le p-1$ such that its residue mod $p$ is a generator of $(\Bbb Z/p\Bbb Z)^\times$. Then $x_j :=$ the residue of $a^{p^{j-1}}$ in $(\Bbb Z/p^j\Bbb Z)^\times$ is a generator of the $C_{p-1}$ we want, and by the fact $(*)$ and iterations of Fermat's little theorem, we also have that the projection $(\Bbb Z/p^j\Bbb Z) \rightarrow (\Bbb Z/p^i\Bbb Z)$ sends $x_j$ to $x_i$.
Finally, for $p=2$, the unit group turns out to be not $C_{2^{j-1}}$, but $C_2 \times C_{2^{j-2}}$ (for $j \ge 3$). To adapt the above, for $y_j$ instead of $1+p$ take $1+p^2$ (i.e. $5$) mod $2^j$, and show its order in the unit group is $2^{j-2}$. Funnily, here the constant $C_2$ part is easy, namely it's generated by $x_j :=$ the residue of $-1$ in $(\Bbb Z/2^j\Bbb Z)$.
Added much later: My above "Everyone since Euler knows ..." was a questionable joke on "Euler's" totient function. In fact, according to Structure of $(\mathbb{Z}/p^k\mathbb{Z})^\times$ other than via $p$-adics?, everyone since Gauß knows how all $(\mathbb Z/p^j)^\times$ (and hence all $(\mathbb Z/n)^\times$) look like -- not just, like Euler, how big they are.
