In Kunen's "The foundation of mathematics" there is an exercise:
Ex: For any infinite cardinal $\kappa$, $|H(\kappa)| = 2^{<\kappa}$.
Prove for $|H(\kappa)| \geq 2^{<\kappa}$ is easy. I have a problem with the second part, $|H(\kappa)| \leq 2^{<\kappa}$.
The proof goes by induction on $\kappa$. For $\kappa = \omega$, $H(\omega) = R(\omega) = \cup_{n<\omega}R(n)$. But $|R(n)| \leq 2^{<\omega}$. Hence $|H(\omega)| \leq 2^{<\omega}\cdot\omega=2^{<\omega}$.
For singular $\kappa$, $\kappa = \cup_{\alpha < cf(\kappa)}\mu_\alpha$, $|H(\kappa)| = |\cup_{\alpha<cf(\kappa)}H(\mu_\alpha)| \leq \cup_{\alpha<cf(\kappa)}2^{<\mu_\alpha}\leq 2^{<\kappa}$.
For successor $\kappa = \theta^+$, we need to prove that $|H(\theta^+)| \leq 2^{\theta}$. We know that $H(\theta)^+ \subseteq R(\theta^+)$. By induction on $\alpha < \theta^+$ prove that $|H(\theta^+) \cap R(\alpha)| \leq 2^\theta$.
For limit $\alpha$, $R(\alpha) = \cup_{\beta<\alpha} R(\beta)$. $$|H(\theta^+) \cap R(\alpha)|=|H(\theta^+) \cap (\cup_{\beta<\alpha} R(\beta))|= |\cup_{\beta<\alpha}(H(\theta^+) \cap R(\beta))| \leq 2^\theta \cdot \alpha = 2^\theta$$
For successor $\alpha=\beta +1$, $H(\theta^+) \cap R(\alpha) = H(\theta^+) \cap P(R(\beta))$.
I do not see why $| H(\theta^+) \cap P(R(\beta))| \leq (2^\theta)^{\theta}$.
Edit: To the comment by Hanul Jeon bellow.
$H(\theta^+) \cap P(R(\beta)) = [H(\theta^+) \cap R(\beta)]^{\leq \theta}$
I think I see why it is so. If $x \in H(\theta^+) \cap P(R(\beta))$ then $|tcl(x)| \leq \theta$ hence $|x| \leq \theta$. $tcl(x) = x \cup \bigcup_{y \in x} tcl(y)$ hence $|tcl(y)| \leq \theta$ for $y \in x$, therefore $y \in H(\theta^+)$ for any $y \in x$ and hence $x \in [H(\theta^+)]^{\leq \theta}$. Since $x \in P(R(\beta))$ so $x \subseteq R(\beta)$ and since $|x| \leq \theta$ then $x \in [R(\beta)]^{\leq \theta}$.
The other side is similar. If $x \in [H(\theta^+) \cap R(\beta)]^{\leq \theta}$ then $|x| \leq \theta$ and $x \subseteq H(\theta^+)$. Since $tcl(x) = x \cup \bigcup_{y \in x} tcl(y)$ and $|tcl(y)| \leq \theta$ for all $y \in x$ then $|tcl(x)| \leq \theta$ hence $x \in H(\theta^+)$. But since $x \subseteq R(\beta)$ so $x \in P(R(\beta))$.
In that case $|H(\theta^+) \cap P(R(\beta))| \leq (2^\theta)^{\theta} = 2^\theta$ that concludes the proof of $|H(\kappa)| = 2^{<\kappa}$.
A very easy way to see why $H(\theta^+)$ injects into $\mathcal P(\theta)$ is to notice that $x\mapsto\operatorname{tcl}(\{x\})$ is an injective function from $H(\theta^+)$ to itself.
Now, by definition $x\in H(\theta^+)$ if and only if $|\operatorname{tcl}(\{x\})|\leq\theta$. So we can code the $\in$ structure of $\operatorname{tcl}(\{x\})$ as a subset of $\theta$.
This, in general, will give you the result immediately for all $\kappa$. Since $x\in H(\kappa)$ if and only if $|\operatorname{tcl}(\{x\})|<\kappa$ if and only if we can code $\operatorname{tcl}(\{x\})$ as a subset of some $\lambda<\kappa$, which gives us an injection from $H(\kappa)$ into $2^{<\kappa}$.