Let $A\colon \operatorname D(A)\to \mathcal H$ be a (generally unbounded but densely defined) self-adjoint operator in a Hilbert space. $$\operatorname D(A^2):=\{\psi \in \operatorname D(A) \text{ s.t. }A\psi \in \operatorname D(A)\}.$$
What is the most direct way to prove that $\operatorname D(A^2)$ is dense in $\mathcal H$? Or it is necessary to use the full machinery of the spectral theorem?
Assume that $A : \mathcal{D}(A)\subseteq X\rightarrow X$ is self-adjoint on the complex Hilbert space $X$. Then $A\pm iI$ are surjective, which also makes the following surjective on $\mathcal{D}(A^2)$: $$ (A+iI)(A-iI)= A^2+I $$ That is enough to imply that $A^2$ is self-adjoint, a result which follows from the theorem given below.
Theorem: Let $B : \mathcal{D}(B)\subset \mathcal{H}\rightarrow\mathcal{H}$ be a symmetric linear operator on a linear domain $\mathcal{D}(B)$. If $B$ is surjective, then $B$ is densely-defined and self-adjoint.
Proof: Suppose that $B$ as described is surjective, and suppose $y\in\mathcal{D}(B^*)$. Then there exists $x\in\mathcal{D}(B)$ such that $Bx=B^*y$, which gives $$ \langle Bz,y\rangle = \langle z,B^*y\rangle=\langle z,Bx\rangle=\langle Bz,x\rangle,\;\;\; z\in\mathcal{D}(B). $$ Therefore, $$ \langle Bz,y-x\rangle = 0,\;\;\; z\in\mathcal{D}(B). $$ It follows that $y=x$ because $B$ is surjective. Hence, $y\in\mathcal{D}(B^*)$ is in $\mathcal{D}(B)$. So $B$ is self-adjoint. $\blacksquare$