Let $\{\mathcal{H}_{n}\}_{n\in \mathbb{N}}$ be a sequence of complex Hilbert spaces and: $$\mathcal{F}(\mathcal{H}) = \bigoplus_{n\in \mathbb{N}}\mathcal{H}_{n}$$ the associated Fock space. Define an operator $N$ on $\mathcal{F}(\mathcal{H})$ as follows: $$\mathcal{F}(\mathcal{F}) \ni \psi = (\psi_{n})_{n\in \mathbb{N}} \mapsto N\psi := (n\psi_{n})_{n\in \mathbb{N}}.$$ This is the so-called number operator, and it is densely-defined on the domain: $$D(N) = \{\psi \in \mathcal{F}(\mathcal{H}): \sum_{n\in \mathbb{N}}n^{2}\|\psi_{n}\|^{2} < \infty\}$$ Because this is a positive operator, we can also consider $\sqrt{N}$ on the associated domain $D(\sqrt{N})$.
Now, consider the following set: $$\mathcal{D}_{0} := \{\psi = (\psi_{n})_{n\in \mathbb{N}}\in \mathcal{F}(\mathcal{H}): \mbox{there exists some $n_{0}$ such that $\psi_{n} = 0$ for every $n > n_{0}$}\}$$ I want to prove that $\mathcal{D}_{0}$ is a core for $\sqrt{N}$. To do that, I will prove that the restriction $\sqrt{N}\bigg{|}_{\mathcal{D}_{0}}$ is essentially self-adjoint so, by uniqueness, $\sqrt{N}$ must be its self-adjoint extension, given that $\mathcal{D}_{0} \subset D(\sqrt{N})$.
My attempt: Suppose that $\psi \in \mathcal{D}_{0}$ is such that $\sqrt{N}\psi = \pm i \psi$. Then, for every $n \in \mathbb{N}$ we have $\sqrt{N}\psi_{n} = \pm i\psi_{n}$, and so $\psi_{0} = 0$ and, for $n \ge 2$, $\|\psi_{n}\|(\sqrt{n}-1) = 0$. Thus, we must have $\psi_{n} = 0$ for $n \ge 2$. Finally, because $\psi_{1} = \pm i \psi_{1}$, we have $\langle \psi_{1}, i\psi_{1}\rangle = i\|\psi_{1}\|^{2} = -i\|\psi_{1}\|^{2} = \langle i\psi_{1},\psi_{1}\rangle$. Thus, $\|\psi_{1}\| = 0$ and $\psi \equiv 0$. In other words, $\text{Ker}(\sqrt{N}\pm i) = \{0\}$ and $\sqrt{N}\bigg{|}_{\mathcal{D}_{0}}$ is essentially self-adjoint by the basic criteria.
Is my proof correct? In particular, does the essentially self-adjointness of $\sqrt{N}\bigg{|}_{\mathcal{D}_{0}}$ really prove that $\mathcal{D}_{0}$ is a core for $\sqrt{N}$? I think this arguments hold, but I am not completely sure. Thanks!