Nitpick : $\Bbb Z_p$ is a sad notation for $p$-adics. Reminds me of cyclic groups. How about $\mathbf{Z}_p$?
The definition is really straightforward as written. $\mathbf{Z}_p$, the set of $p$-adic integers, is the collection of "infinite-tuples" $(x_0, x_1, x_2, \cdots )$ with $x_k \in \Bbb Z/p^{k+1}\Bbb Z$ such that each $x_{i+1}$ is equivalent to $x_i$ modulo $p^{i+1}$.
Wait, wait, that didn't even made sense to me(!). I have a great ignorance for these kind of symbolic definitions, so they won't make sense to anyone. Let's look at examples, they always give a good notion of what is going on "behind the scene".
Consider the equation $x^2 = -1$ in $\Bbb Z$. This has, of course, no solutions. However, this has solutions when considered over certain finite fields. $x^2 = -1$ has trivial solutions over $\Bbb Z/Z\Bbb Z$ (itself a trivial group :p), no solutions over $\Bbb Z/3\Bbb Z$ (only quadratic res. are 1 and 0), neither any over $\Bbb Z/4\Bbb Z$ (only quadratic res. again are 1 and 0). The smallest field in which it does is $\Bbb Z/5\Bbb Z$ : $2^2 = 4 = -1$ and $3^2 = 9 = 10 - 1 = -1$.
Consider the 2 modulo 5 solution. So we know that $x = 2 + 5y$ and we can sub that in : $-1 = x^2 = (2 + 5y)^2 = 4 + 25y^2 + 20y$ which reduces to $20y = -5 + 25y^2$, i.e., $20y = -5 \pmod{25}$ which after inversion gives $y = 1 \pmod{5}$. We can sub that back in $x = 2 + 5y$ to get $x = 2 + 5(1 + 5y_1)$ thus $x = 7 \pmod{25}$, which is in turn a solution in $\Bbb Z_{25}$. By similar calculations, you'll end up with $x = 57 \pmod{125}$ in $\Bbb Z/{125}\Bbb Z$ and so on and so on.
The specialty of this chain of solutions is once found a solution $x_{i+1}$ in $\Bbb Z_{5^{i+1}}$, $x_{i+1}$ can be mapped back to $x_i$ modulo $5^i$ (that's precisely the way they are constructed). Thus you can construct an "infinite-tuple" $x = (2, 7, 57, \cdots)$ in the infinite direct product $\prod \Bbb Z/5^{i+1} \Bbb Z$ such that each elt $x_{i+1}$ can be "mapped back" to $x_i$ using homomorphisms (in this case, modulo $5^{i+1}$). This construction precisely gives us a ring, and out "infinite tuple" satisfies $x^2 = -1$ so we are kind of "expanding out" the usual integers. This is a possible construction of $5$-adic integers, and in formal terms, your symbolic definition with $n = 5$ subbed in.
This is all there is to the definition of $p$-adics. They are the collection of a bunch of elements from $\Bbb Z/p^k \Bbb Z$ for $k > 0$ such that there is a collection of homomorphisms from $\Bbb Z/p^{i+1}\Bbb Z$ to $\Bbb Z/p^i \Bbb Z$. The notion of inverse limits are just a generalization of this idea.
EDIT Thought that I'd rather add this to make the concept clear. Consider $10$-adic numbers (warning - not a field). Elements are all of the form $(x_1, x_2, x_3, \cdots )$ such that $x_i \in \Bbb Z/10^i \Bbb Z$ for all $i$. Recall that an integer modulo $10^i$ is really the first $10^i$ digits of the integer, the rest of them chopped off. In that case, for $(x_1, x_2, \cdots )$, $x_1$ can be viewed as the last digit of some integer, $(x_2 - x_1)/10$ as the second digit from the right, $(x_3-x_2)/100$ as the third digit from the right, and so on and so forth. Thus, every $10$-adic $(x_i)$ has a corresponding infinite series representation $\sum_{n = 0}^\infty a_n 10^n$. For in general $k$-adic numbers, similar can be stated, but the "infinite integers" then becomes "infinite integers base $k$", i.e., the series representation is of the form $\sum_{n = 0}^\infty a_n k^n$. From this you can verify that every integer is a $p$-adic integer for some prime $p$ : take $13$ and $p = 3$. Modulo $3$, $13 = 1$ hence our sequence is $(1, \cdots)$. $13 = 4$ modulo $9$, thus we get another term $(1, 4, \cdots)$. However, all the higher powers of $3$ are greater than $13$ so the rest of the terms are all $13$ : the desired $3$-adic expansion is $(1, 4, 13, 13, 13, \cdots)$. However, there are $p$-adics that can't be represented as integers - this is pretty obvious to construct as every $p$-adic rep for integers eventually stabilize to some fixed integer.
