Intuition for $\omega^\omega$

I'm trying to understand the ordinal number $\omega^\omega$ and I'm having a hard time. I think I understand what $\omega^2$ is. It's what I would get if I took countably many copies of $\omega$ and order the set of those copies like $\omega.$ $\omega^3$ is a bit more difficult because I need to take countably many copies of $\omega^2$ now and do the same with them. I understand that the next ones are made in the same way, recursively. But understanding that the definition has meaning doesn't help me very much with visualizing $\omega^{10}$. And I become completely helpless when I try to tell myself what $\omega^\omega$ is in plain words. Is some explanation possible that is not just the definition (which I know)?

I think of $\omega^n$ as the set of all $n$-element sequences of naturals (not just increasing ones as in Andres's answer), ordered lexicographically.

There are two possible ways to extend this image to $\omega^\omega$. One is to say that $\omega^\omega$ is the set of all finite sequences of naturals, ordered with shorter sequences first, and them lexicographically between sequences of the same length.

The other is to say that an element of $\omega^\omega$ is an "infinite-to-the-left" sequence $(\ldots,a_n,\ldots,a_3,a_2,a_1,a_0)$ such that only finitely many elements of the sequence are nonzero. These sequences are then ordered lexicographically.

These "infinite-to-the-left" sequences perhaps look less strange if we write them as $$ \cdots + a_nX^n + \cdots + a_3X^3+a_2X^2+a_1X + a_0 $$ that is, they are just the set of formal polynomials with coefficients in $\mathbb N$. (The usual definition of "formal polynomial" takes care of the requirement that cofinitely many of the elements must be zero).

We can even write $\omega$ instead of the formal variable $X$: $$ \cdots + \omega^n\cdot a_n + \cdots + \omega^3\cdot a_3+\omega^2\cdot a_2+ \omega \cdot a_1 + a_0 $$ and written this way, the elements are exactly the actual ordinals below $\omega^\omega$. Because of the non-commutative nature of ordinal multiplication, we have to write the coefficients to the right of the power of $\omega$, though.

(Be careful not to take this too far, however: the usual rules for adding and multiplying polynomials do not correspond to ordinal addition and multiplication).

Here's how I visualize it. Consider this subset of $[0,1]$:

$$E_1 \;\; = \;\; \left\{\frac{1}{2}, \;\; \frac{2}{3}, \;\; \frac{3}{4}, \;\; \frac{4}{5}, \;\; \frac{5}{6}, \;\; \frac{6}{7}, \;\; \ldots \right \}$$

This is $\omega$.

More precisely, when ordered by the usual ordering for real numbers, $E_1$ becomes a representation for the order type $\omega.$

Now consider the union of $E_1$ and its $1$-unit translate (a subset of $[1,2]$), which will be a subset of $[0,2]$.

This is $\omega 2$.

By taking the unions of various appropriate integer translates of $E_1$, we get $\omega 3, \; \omega 4, \; \omega 5, \; \ldots$

By putting an integer translate of $E_1$ into the interval $[n, \; n+1],$ for each non-negative integer $n,$ and taking the union of these countable many translates of $E_1,$ we get $\omega \omega = {\omega}^2.$

If we take this last set, the set representing ${\omega}^2$, and scale/translate it into the interval $[0,1]$ (do this in a way the preserves order, which will be the case if you use a positive scale factor), we'll have a representation of ${\omega}^2$ that is a subset of $[0,1].$ Call this last set $E_2.$

The union of $E_2$ with its $1$-translate will be ${\omega}^2 2$.

By taking the unions of various appropriate integer translates of $E_2$, we get ${\omega}^2 3, \; {\omega}^2 4, \; {\omega}^2 5, \; \ldots$

You can also get things like ${\omega}^2 3 + \omega 2 + 4$ by taking the union of a set in $[0,3]$ that represents ${\omega}^2 3$ with a set in $[4,5]$ that represents $\omega$ with a set in $[5,6]$ that represents $\omega$ with the set $\left\{6 + \frac{1}{2}, \; 6 + \frac{2}{3}, \; 6 + \frac{3}{4}, \; 6 + \frac{4}{5} \right\}.$

By putting an integer translate of $E_2$ into the interval $[n, \; n+1],$ for each non-negative integer $n,$ and taking the union of these countable many translates of $E_2,$ we get ${\omega}^2 \omega = {\omega}^3.$

If we take this last set, the set representing ${\omega}^3$, and scale/translate it into the interval $[0,1]$ (do this in a way the preserves order, which will be the case if you use a positive scale factor), we'll have a representation of ${\omega}^3$ that is a subset of $[0,1].$ Call this last set $E_3.$

Keep going in this manner, obtaining $E_4 \subseteq [0,1]$ that is a representation of ${\omega}^4,$ $E_5 \subseteq [0,1]$ that is a representation of ${\omega}^5,$ $E_6 \subseteq [0,1]$ that is a representation of ${\omega}^6, \; \ldots$

Then ${\omega}^{\omega}$ can be represented by

$$ E_1 \; \cup \; \left(E_2 + 1 \right) \; \cup \; \left(E_3 + 2 \right)\; \cup \; \left(E_4 + 3 \right)\; \cup \; \left(E_5 + 4 \right) \; \cup \; \ldots,$$

where $E_{n+1} + n$ represents the $n$-translate of $E_{n+1}$ for $n = 1, \; 2, \; 3, \; \ldots$