Could someone explain aleph numbers? I am having trouble understanding aleph numbers. I understand $\aleph_0$ is a countable infinity, but after that, I'm lost. What are $\aleph_1,\aleph_2,\aleph_3$, etc. to $\aleph_n$? Is there an infinite amount of $\aleph$ numbers or do they stop at some finite number? 

This is an issue about which mathematicians who haven't studied set theory beyond what they actually use are often confused.
$\aleph_1$ is the cardinality of the set of all countable ordinal numbers.
$\aleph_2$ is the cardinality of the set of all ordinal numbers of cardinality $\le\aleph_1$.
$\aleph_3$ is the cardinality of the set of all ordinal numbers of cardinality $\le\aleph_2$.
${}\qquad\vdots$
$\aleph_\omega$ is the cardinality of the set of all ordinal numbers of cardinality no bigger than some $\aleph_n$ for finite $n$.
$\aleph_{\omega+1}$ is the cardinality of the set of all ordinal numbers of cardinality $\aleph_\omega$.
${}\qquad\vdots$
There exists an $\aleph_\alpha$ for every ordinal number $\alpha$.
If every set can be well-ordered --- a proposition equivalent to the axiom of choice --- then every infinite cardinality is one of these.
$2^{\aleph_0}$ is the cardinality of the set of all subsets of a set of cardinality $\aleph_0$.  If the continuum hypothesis is true, then there is no cardinality between $\aleph_0$ and $2^{\aleph_0}$.  If every cardinality is one of the alephs, then the continuum hypothesis is equivalent to $2^{\aleph_0}=\aleph_1$.  It is consistent with the Zermelo--Fraenkel axioms plus the axiom of choice that $2^{\aleph_0} = \aleph_n$ for some finite $n>1$.  It is not consistent that $2^{\aleph_0} = \aleph_\omega$ since it can be shown that $2^{\aleph_0}$ is not the supremum of any countable set of cardinalities.  But it is consistent that $2^{\aleph_0} >\aleph_\omega$.
