Suppose $$A$$ is a densely defined closed operator. Show that $$A^*A$$ (with domain $$D(A^*A)=\{\psi\in D(A)|A\psi\in D(A^*)\}$$) is self-adjoint.

Let $$\psi\in D(A^*A)$$ and observe that $$A^*A$$ is nonnegative: $$(\psi,A^*A\psi)=(A\psi,A\psi)=\|A\psi\|\ge0.$$ Then $$A^*A$$ is symmetric and $$(A^*A\psi,\psi)=(\psi,A^*A\psi)=((A^*A)^*\psi,\psi)$$ for all $$\psi\in D(A^*A)\cap D((A^*A)^*)$$, implying $$(A^*A)^*=A^*A$$ in this domain. We must show that $$D(A^*A)=D((A^*A)^*)$$. Recall the definition of the domain of the adjoint: $$D(B^*)=\{\psi\in\mathcal{H}|\exists\tilde\psi\in\mathcal{H}:(\psi,B\phi)=(\tilde\psi,\phi),\forall\phi\in D(B)\}.$$ Then for a symmetric operator, say $$B$$, we have $$D(B)\subset D(B^*)$$ since $$(\psi,B\phi)=(B\psi,\phi)$$ for all $$\phi\in D(B)$$. This gives us the forward inclusion since $$A^*A$$ is symmetric. How can the reverse inclusion be achieved?

The graph of $$\mathcal{G}(A)$$ is closed. The orthogonal complement of $$\mathcal{G}(A)$$ consists of all pairs $$(y,z)\in\mathcal{H}\times\mathcal{H}$$ such that $$\langle (x,Ax),(y,z)\rangle =0,\;\; x\in\mathcal{D}(A).$$ That is the same as $$\langle x,y\rangle+\langle Ax,z\rangle=0,\;\; x\in\mathcal{D}(A),$$ which is exactly the requirement that $$z\in\mathcal{D}(A^*)$$ and $$A^*z=-y$$. Therefore, every $$(z,w)\in\mathcal{H}\times\mathcal{H}$$ may be written as the following sum for some $$x\in\mathcal{D}(A)$$ and $$y\in\mathcal{D}(A^*)$$: $$(z,w) = (x,Ax)+(-A^*y,y)$$ Therefore, $$z=x-A^*y,\;\; w=Ax+y$$ We can choose $$w=0$$ in order to obtain $$x,y$$ such that $$z=x-A^*y=x+A^*Ax.$$ It follows that $$A^*A+I$$ is surjective because for every $$z$$ there exists $$x$$ such that $$z=(I+A^*A)x$$. It is easy to verify that $$A^*A+I$$ is symmetric on its natural domain.
The domain of $$A^*A+I$$ has to be dense because it is surjective; to see why, suppose $$y\perp\mathcal{D}(A^*A)$$. Then there exists $$x\in\mathcal{D}(I+A^*A)$$ such that $$(I+A^*A)x=y$$. Because $$y\perp x$$, it follows that $$0= \langle y,x\rangle = \langle (I+A^*A)x,x\rangle=\|x\|^2+\|Ax\|^2\implies x=0 \implies y = 0.$$ So $$I+A^*A$$ is densely-defined, symmetric and surjective.
That is enough to conclude that $$I+A^*A$$ is self-adjoint. Indeed, suppose $$y\in\mathcal{D}((I+A^*A)^*)$$. Then $$\langle (I+A^*A)x,y\rangle = \langle x,(I+A^*A)^*y\rangle,\;\; x\in\mathcal{D}(I+A^*A).$$ Then there exists $$w\in\mathcal{D}(I+A^*A)$$ such that $$(I+A^*A)^*y=(I+A^*A)w$$. Hence, $$\langle (I+A^*A)x,y\rangle = \langle x,(I+A^*A)^*y\rangle= \langle x,(I+A^*A)w\rangle = \langle (I+A^*A)x,w\rangle$$ Because $$I+A^*A$$ is surjective, then $$y=w$$, from which it follows that $$y\in\mathcal{D}(I+A^*A)$$ and $$(I+A^*A)^*y=(I+A^*A)y$$. Therefore, $$(I+A^*A)^*=I+A^*A$$.