Let $\mathscr{F}$ be the set of all functions of the form $f: \mathbb{N}^k \to \mathbb{N}$ for some $ k \ge 1$.
Let $G$ be the set of all partial recursive functions; that is, $G = \lbrace f \in \mathscr{F} | f \text{ is partial recursive}\rbrace$. What is the cardinality of $\mathscr{F}$? What is the cardinality of $G$?
I'm not sure where to start with this one, although my intuition for cardinality of $G = \aleph_0 $ and cardinality of $\mathscr{F} = \aleph_k$.
Any help proving these or correcting my intuition would be appreciated
I assume you mean $k\geq 1$, not $k\leq 1$.
The cardinality of the set of all functions from $A$ to $B$ has cardinality $|B|^{|A|}$ (that's why the notation for that set is $B^A$).
Since $\mathbb{N}^k$ has cardinality $\aleph_0$, the cardinality of $\mathscr{F}$ is $|\mathbb{N}|^{|\mathbb{N}^k|} = \aleph_0^{\aleph_0} = 2^{\aleph_0}$. For the latter equality, simply note that $$2^{\aleph_0} \leq \aleph_0^{\aleph_0}\leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\aleph_0} = 2^{\aleph_0}.$$
It could equal $\aleph_r$ for some $r\in\omega$, $r\geq 1$ (in fact, for each such $r$ there are set theories in which it equals exactly that), by results of Cohen, but we don't know what it is equal to; the exact aleph that equals $2^{\aleph_0}$ is undecidable in standard set theory. It does not, however, depend on the value of $k$.
As for the set of partial recursive functions, the set of partial recursive functions from $\mathbb{N}$ to itself is $\aleph_0$ (the set is countable), hence so is $G$.