Largest eigenvalues of AA' equals to A'A I need help with proving that for any real matrix,the largest eigenvalue of AA' equals to the largest eigenvalue of A'A

For any $m \times n$ and $n \times m$ matrices $A$ and $B$, the nonzero eigenvalues of $AB$ and $BA$ are the same. Namely, if $AB v = \lambda v$ with $\lambda \ne 0$ and $v \ne 0$, then $Bv \ne 0$ and $BA(Bv) = B(AB v) = \lambda B v$.
