If $\beth_0$ is the cardinality of the natural numbers, $\beth_1$ the cardinality of the continuum, $\beth_2$ the cardinality of the set of all functions from $\mathbb{R}$ to $\mathbb{R}$, where $\mathbb{R}$ is the set of Real numbers, then is the cardinality of the set of sequences of functions from $\mathbb{R}$ to $\mathbb{R}$, $\beth_1$?
Thanks in advance !
If I understand this correctly, you are asking what is $(\beth_2)^{\beth_0}$, which is the cardinality of the set $\left(\Bbb{R^R}\right)^\Bbb N$.
This is easier to think of in terms of repeated exponentiation, and in terms of $\aleph$'s. Remember $\aleph_0=\beth_0$.
$$(\beth_2)^{\beth_0}=\left(2^{2^{\aleph_0}}\right)^{\aleph_0}=2^{2^{\aleph_0}\cdot\aleph_0}=2^{2^{\aleph_0}}=\beth_2.$$
Generally, if a set has size $\kappa$, then $\kappa\leq\kappa^\lambda$, whenever $\lambda>0$. So if we start with a set of size $\beth_2$ it is impossible that by raising it to the power of $\aleph_0$ we get $\beth_1$.