# What is $\aleph_0$ powered to $\aleph_0$?

By definition $\aleph_1 = 2 ^{\aleph_0}$. And since $2 < \aleph_0$, then $2^{\aleph_0} = {\aleph_1} \le \aleph_0 ^ {\aleph_0}$. However, I do not know what exactly $\aleph_0 ^ {\aleph_0}$ is or how I could compute it.

No. By definition $\aleph_1$ is the least uncountable $\aleph$ number. $2^{\aleph_0}$ can be quite a large $\aleph$, or it could be $\aleph_1$. For example, many forcing axioms (e.g. the proper forcing axiom) prove that $2^{\aleph_0}=\aleph_2$.
The assertion $2^{\aleph_0}=\aleph_1$ is known as The Continuum Hypothesis and was proven unprovable from the usual axioms of set theory. We can therefore add axioms which decide the continuum hypothesis, e.g. itself or the aforementioned forcing axiom.
$$2^{\aleph_0}\leq\aleph_0^{\aleph_0}\leq (2^{\aleph_0})^{\aleph_0}= 2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}$$