Showing properties of Moore–Penrose matrix using the SVD decomposition

Question

Using the Singular Value Decomposition, show:

1. $\operatorname{Rank}(A)=\operatorname{Rank}(AA^*)=\operatorname{Rank}(A^*A)$.
2. $A^{+}=A^*(AA^*)^{-1}$.
3. $\operatorname{null}(A^{+})=\operatorname{null}(A^*)$.

where $A^+$ is the Moore-Penrose matrix, and $A^*$ is the transpose conjugate matrix.

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There are many identities on the Moore-Penrose matrix, on the following Wikipedia page, with no proofs.

Moore-penrose matrix: https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse.

Part 1 can be shown with the rank-theorem since $\operatorname{null}(A)=\operatorname{null}(AA^*)$ thus $\dim(\operatorname{null}(A))=\dim(\operatorname{null}(A^*))$ which is the same as, $\operatorname{rank}(A)=\operatorname{rank}(AA^*)$. And if we use SVD decomposition it will give us the same result:

$A=VDU^*$ where $V$ and $U$ are orthogonal matrix, and $D=diag{\sigma_1\ldots\sigma_n}$. Where $\sigma_1,\ldots,\sigma_n$ are the singular values of $A$. It is easy to see that $Rank(A)=Rank(D)$. Then,

$A^*=UD^*V^*$.

$AA^*=V(DD^*)V^*$

$A^*A=U(D^*D)U^*$

So, we get the SVD of $AA^*$ and $A^*A$.

It is obvious that $rank(DD^*)=rank(D^*D)=rank(D)$. Thus, $rank(AA^*)=rank(A^*A)=rank(A)$

But I did not manage to use the SVD to show part 2 and part 3 in the properties above. Any help would be welcomed.