Do $\omega^\omega=2^{\aleph_0}=\aleph_1$? As we know, $2^{\aleph_0}$ is a cardinal number, so it is a limit ordinal number. However, it must not be $2^\omega$, since $2^\omega=\sup\{2^\alpha|\alpha<\omega\}=\omega=\aleph_0<2^{\aleph_0}$, and even not be $\sum_{i = n<\omega}^{0}\omega^i\cdot a_i$ where $\forall i \le n[a_i \in \omega]$. Since $\|\sum_{i = n<\omega}^{0}\omega^i\cdot a_i\| \le \aleph_0$ for all of them.
Besides, $\sup\{\sum_{i = n<\omega}^{0}\omega^i\cdot a_i|\forall i \le n(a_i \in \omega)\}=\omega^\omega$, and $\|\omega^\omega\|=2^{\aleph_0}$ since every element in there can be wrote as $\sum_{i = n<\omega}^{0}\omega^i\cdot a_i$ where $\forall i \le n[a_i \in \omega]$ and actually $\aleph_{0}^{\aleph_0}=2^{\aleph_0}$ many. 
Therefore $\omega^\omega$ is the least ordinal number such that has cardinality $2^{\aleph_0}$, and all ordinal numbers below it has at most cardinality $\aleph_0$. Hence $\omega^\omega=2^{\aleph_0}=\aleph_1$? 
 A: There are very different definitions for cardinal and ordinal exponentiation. Ordinal exponentiation is defined in a way which allows us to generate well-orderings of a particular set; where as cardinal exponentiation strips out the ordering and deals with cardinality of all functions from one set to another.
This is why some authors differentiate the two by using $^\omega\omega$ for cardinal exponentiation and $\omega^\omega$ for ordinal exponentiation (at least where context is ambiguous). Personally I am not a big fan of this approach, despite the fact it may clear some possible confusion.
Lastly, as commented,  $2^{\aleph_0}$ need not be equal to $\aleph_1$. This is known as The Continuum Hypothesis which was proved unprovable from ZFC. 
To add on the confusion, let me give a short list of some common uses for $\omega^\omega$:


*

*The first limit ordinal which is a limit of limit ordinals each a limit of limit ordinals which are not limits of limit ordinals.

*The set of all sequences of natural numbers (which also form the underlying set for the following uses).

*The Baire space.

*The real numbers (in some contexts).
A: Your notation confuses cardinal and ordinal exponentiation, which are two very different things. If you’re doing cardinal exponentiation, $2^\omega$ is exactly the same thing as $2^{\aleph_0}$, just expressed in a different notation, because $\omega=\aleph_0$. If you’re doing ordinal exponentiation, then as you say, $2^\omega=\omega$.
But if you’re doing ordinal exponentiation, then $$\omega^\omega=\sup_{n\in\omega}\omega^n=\bigcup_{n\in\omega}\omega^n\;,$$ which is a countable union of countable sets and is therefore still countable; it doesn’t begin to reach $\omega_1$. Similarly, still with ordinal exponentiation, $\omega^{\omega^\omega}$ is countable, $\omega^{\omega^{\omega^\omega}}$ is countable, and so on. The limit of these ordinals, known as $\epsilon_0$, is again countable, being the limit of a countable sequence of countable ordinals, and so is smaller than $\omega_1$. (It’s the smallest ordinal $\epsilon$ such that $\omega^\epsilon=\epsilon$.)
Now back to cardinal exponentiation: for that operation you have $2^\omega\le\omega^\omega\le(2^\omega)^\omega=2^{\omega\cdot\omega}=2^\omega$, where $\omega\cdot\omega$ in the exponent is cardinal multiplication, and therefore $2^\omega=\omega^\omega$ by the Cantor-Schröder-Bernstein theorem. The statement that this ordinal is equal to $\omega_1$ is known as the continuum hypothesis; it is both consistent with and independent of the other axioms of set theory.
