Why does $\operatorname{rank}([u_{n-k+1},...,u_n][v_{n-k+1},...,v_n]^*)=k$? Let $A\in M_n$ and  the (nonzero) minimum singular 
value of the matrix $A$ has multiplicity $k\ge1$,
$${s_1} \ge {s_2} \ge .... \ge {s_{n - k}} > {s_{n - k + 1}} = .... = {s_n} > 0$$         .
with associated left singular vectors $u_1,u_2,....u_n$ and associated right singular vectors $v_1,v_2,....v_n$ .
List item
It is true that           .
$rank([u_{n-k+1},...,u_n][v_{n-k+1},...,v_n]^*)$=k,  and 
$rank([u_1,...,u_n][v_1,...,v_n]^*)$=n?
 A: Let $\sigma$ be a singular value of $A$ with respective left- and right singular vectors $u$ and $v$. We can assume that $\|v\|=1$. Then also $\|u\|^2 = \langle u,u\rangle = \sigma^{-1}\langle Av,u\rangle = \sigma^{-1}\langle v,A^*u\rangle = \langle v,v\rangle = 1$. We have $Av = \sigma u$ and $A^*u = \sigma v$. Hence, $A^*Av = \sigma A^*u = \sigma^2 v$ and $AA^*u = \sigma^2 u$. Hence $u$ and $v$ are eigenvectors of $AA^*$ and $A^*A$, respectively, corresponding to the eigenvalue $\sigma^2$. This means that for different singular values $\sigma_1$ and $\sigma_2$ we have that $v_1\perp v_2$ and $u_1\perp u_2$. Hence, in the singular value decomposition $U$ and $V$ are unitary matrices.
However, in the original post we have that the smallest singular value is of multiplicity $k$ and nothing is said about whether the corresponding $u_j$'s are orthogonal to each other or only linearly independent (same for the $v_j$'s). So, in principle, we could have $u_{n-k+1} = \ldots = u_n$ (same for the $v_j$'s). Then, of course, both statements are false. They are true if the vectors are chosen to form two ONBs (which they are in the SVD). In summary, the question is undecidable here because there is information missing.
