How does one compute the cardinality of the set of functions $f:\mathbb{R} \to \mathbb{R}$ (not necessarily continuous)?
All you need is a few basics of cardinal arithmetic: if $\kappa$ and $\lambda$ are cardinals, none of them zero, and at least one of them is infinite, then $\kappa+\lambda = \kappa\lambda = \max\{\kappa,\lambda\}$. And cardinal exponentiation satisfies some of the same laws as regular exponentiation; in particular, $(\kappa^{\lambda})^{\nu} = \kappa^{\lambda\nu}$. 

The cardinality of the set of all real functions is then
$$|\mathbb{R}|^{|\mathbb{R}|} =\mathfrak{c}^{\mathfrak{c}} = (2^{\aleph_0})^{2^{\aleph_0}} = 2^{\aleph_02^{\aleph_0}} = 2^{2^{\aleph_0}} = 2^{\mathfrak{c}}.$$
In other words, it is equal to the cardinality of the power set of $\mathbb{R}$.

With a few extra facts, you can get more. In general, if $\kappa$ is an infinite cardinal, and $2\leq\lambda\leq\kappa$, then $\lambda^{\kappa}=2^{\kappa}$. This follows because:
$$2^{\kappa} \leq \lambda^{\kappa} \leq (2^{\lambda})^{\kappa} = 2^{\lambda\kappa} = 2^{\kappa},$$
so you get equality throughout. The extra information you need for this is to know that if $\kappa$, $\lambda$, and $\nu$ are nonzero cardinals, $\kappa\leq\lambda$, then $\kappa^{\nu}\leq \lambda^{\nu}$.

In particular, for any infinite cardinal $\kappa$ you have $\kappa^{\kappa} = 2^{\kappa}$. 
