Assuming the Axiom of Choice: ${\mathrm{\aleph }}_0^{{\mathrm{\aleph }}_0}=2^{{\mathrm{\aleph }}_0}$ and $\mid {\mathbb{R}}^{\mathbb{R}}\mid =2^{2^{{\mathrm{\aleph }}_0}}$, computed from the exponent laws and Hessenberg

Assume the Axiom of Choice (The Axiom of Choice). Write $\mathfrak{c}=2^{{\mathrm{\aleph }}_0}$, and let ${\mathbb{R}}^{\mathbb{R}}$ denote the set ${}^{\mathbb{R}}\mathbb{R}$ of all functions $\mathbb{R}\to \mathbb{R}$, continuity playing no role. Then

${\mathrm{\aleph }}_0^{{\mathrm{\aleph }}_0}=2^{{\mathrm{\aleph }}_0}=\mathfrak{c},\mid {\mathbb{R}}^{\mathbb{R}}\mid =2^{2^{{\mathrm{\aleph }}_0}}\mathrm{.}$

Both computations are squeezes: an upper and a lower bound that meet, with Hessenberg: $\kappa \otimes \kappa =\kappa$ for every infinite cardinal $\kappa$, proved in ZF from the canonical well-order of $\kappa \times \kappa$ closing the gap through $\kappa \otimes \kappa =\kappa$ and the second exponent law (Commutativity, associativity, distributivity and monotonicity of $\oplus$ and $\otimes$, the unit laws, the two exponent laws, and $\kappa \le \lambda$ if and only if $\kappa$ injects into $\lambda$) turning a repeated exponent into a product.

The second value is worth reading against the first. There are $\mathfrak{c}$ real numbers and $2^{\mathfrak{c}}$ functions between them, so the set of all real functions is strictly larger than the continuum (Assuming the Axiom of Choice, $2^{\kappa }=\mid \mathcal{P}(\kappa )\mid$, and Cantor's theorem in cardinal form: $\kappa <2^{\kappa }$), by exactly one application of the power operation.

Why the first computation is a collapse and not a coincidence. Any base between $2$ and $2^{{\mathrm{\aleph }}_0}$ gives the same value when raised to ${\mathrm{\aleph }}_0$, because the chain of step 1.1 closes on both sides. That is the general phenomenon recorded in FALSE: $\kappa <\lambda$ implies ${\kappa }^{\mu }<{\lambda }^{\mu }$: strict monotonicity in the base is false, and this is the smallest instance.

Where Hessenberg's theorem enters. Twice, both times as $\kappa \otimes \kappa =\kappa$ turning a repeated exponent into a single one: at ${\mathrm{\aleph }}_0$ in step 1.1 and, through absorption, at $\mathfrak{c}$ in step 2.1. Without it neither exponent could be simplified and both computations would stall at an upper bound.

Continuity is irrelevant here, and that is worth saying. ${\mathbb{R}}^{\mathbb{R}}$ above is the set of all functions, with no regularity assumed. Counting the continuous ones is a different computation, needing tools this page and the pages it rests on do not provide, and no claim about it is made here.