SVD uniquely determined for a matrix

Question

I'm trying to prove the following:

1. Given a matrix, the singular values of that matrix are uniquely determined.
2. If that matrix is square, and the singular values are distinct, the left and right singular vectors are uniquely determined up to a sign.

These are suppose to be corollaries of the relation between SVD's and eigenvalues decomposition, i.e. that if $A = U\Sigma V'$, then $A'A = V\Sigma^2 V'$.