The reason is that the definitions of ordinal arithmetics and cardinal arithmetics are very different.
Whereas the ordinal arithmetics operations are concerned with order types, the cardinal arithmetics are concerned with certain sets.
For example, $\alpha+\beta$ as ordinals is the order type of $\alpha$ concatenated with $\beta$. Whereas cardinality strips naked any possible structure, and considering the cardinality of the set $\{0\}\times\alpha\cup\{1\}\times\beta$, which is equal to the maximum of $|\alpha|$ and $|\beta|$ (granted one is infinite).
Exponentiation, which is the strangest one, is defined very differently, again, from ordinals and cardinals.
- In cardinals $\alpha^\beta$ is the cardinality of the set of all functions from $\beta$ to $\alpha$.
In the ordinals, we care about the order, so $\alpha^\beta$ is the order type of the reverse lexicographic order of functions from $\beta$ into $\alpha$ which are non-zero only in finitely many coordinates.
Equivalently, and perhaps more clearly, we can define this by induction, $\alpha^0=1$, $\alpha^{\beta+1}=\alpha^\beta\cdot\alpha$, and for a limit ordinal $\delta$, $\alpha^\delta=\sup\{\alpha^\beta\mid\beta<\delta\}$.
Now we easily see that $2^\omega=\omega$ when we are talking about ordinal exponentiation. $2^\omega$ is the limit of $2^n$ for finite $n$, but $2^n$ has a finite order type, and it's a strictly increasing sequence. The limit of a strictly increasing sequence of finite ordinals is $\omega$ itself.
Well, it sounds weird, isn't it? But it's not really weird. We have this sort of phenomenon in other - more familiar - systems of arithmetics.
In the natural numbers $n\cdot m$ can be defined as repeated addition of $n$, $m$ times. Addition itself can be defined as repeated successor operation. On the other hand, when we consider the real numbers $\sqrt2\cdot\sqrt2$ cannot be thought as repeated addition. What does it even mean to add something $\sqrt2$ times? It is true that in this case, if we restrict back to the natural numbers then the operations become repeated application of the "previous operation", but this is because of the nature of the natural numbers as a corner stone of modern mathematics (in many many ways). For infinite, and less-corner stoney this is not the case, as exhibited in ordinals and cardinals.
$\omega + 1$ is a good example. $\omega + 1$ is the ordinal $\{0, 1, 2, \ldots, \omega\}$; as an ordering, think of it as $\omega$ with one more element at the end. It is greater than $\omega$, because it has $\omega$ as a proper initial segment; for ordinals, that's what "greater" means. But it's not bigger than $\omega$ - there's a bijection between $\omega$ and $\omega + 1$, given by the function $f$ which takes $0$ to $\omega$ and $n$ to $n - 1$ for all $n > 0$.
By contrast, the ordinal $\omega_1$, defined as the first uncountable ordinal, is bigger than all of its predecessors, by definition - if $\alpha < \omega_1$, then $\alpha$ can't be uncountable, so there is an injection from $\alpha$ to $\omega$. But there's no injection from $\omega_1$ to $\omega$, by definition, because that would make $\omega_1$ countable. So $\omega_1$ is a cardinal, often denoted $\aleph_1$.
The key idea here is that the author is using "bigger" to refer to size, not ordering - that is, "bigger" is a statement about whether a certain injection exists, not where an ordinal appears in the standard ordering.
On the other hand, you asked about an ordinal that is "equal to or smaller than some of its predecessors". This can't happen. An ordinal is never equal to its predecessors, because different ordinals are always different - it's like asking whether there's a number that's equal to a different number. And since every ordinal has an injection into all of its successors, ordinals can't decrease in size. The only thing that can happen is the example I've outlined above, where we have an ordinal that's the same size as its predecessor. Note that this means that, for ordinals, "the same size as" and "equal to" do not mean the same thing.
Best Answer
No. Both $\omega$ and $\omega + 1 = \omega \cup \{\omega\}$ are equinumerous to $\omega$, but they are not the same.
Because it's convenient and always exists. There isn't a second-minimum ordinal equinumerous to $7$, for instance. On the other hand, there isn't a maximum ordinal equinumerous to $\omega$ (all ordinals below $\omega_1$ are equinumerous to it). So what other choice do you propose?
It's the smallest ordinal that has the same size as $A$. That's it.