AAáµ€ and Aáµ€A have the same characteristic polynomial

Idea. The characteristic polynomial is $\chi_{M}(t)=\det(tI-M)$. We compare $\det(tI-XY)$ and $\det(tI-YX)$ for $t\ne0$ using the identity $\det(I+MN)=\det(I+NM)$, then conclude since two polynomials that agree for infinitely many $t$ are equal.

Step 1: $t\ne0$. Write
$$\det(tI-XY)=t^n\det\!\left(I-\frac Xt\,Y\right)=t^n\det\!\left(I-Y\,\frac Xt\right)=\det(tI-YX),$$
where the middle equality is the identity $\det(I+MN)=\det(I+NM)$ with $M=-X/t$ and $N=Y$.

Step 2: all $t$. $\chi_{XY}$ and $\chi_{YX}$ are polynomials in $t$ which coincide for all $t\ne0$, i.e. at infinitely many points, so they are equal as polynomials.

Step 3: application. Take $X=A$, $Y=A^T$. Then $\chi_{AA^T}=\chi_{A^TA}$. Both matrices are symmetric, hence diagonalisable with real eigenvalues; the eigenvalues counted with algebraic multiplicity are the roots of the characteristic polynomial, and for symmetric matrices algebraic and geometric multiplicity coincide. So the spectra agree with multiplicity.

Remark. For rectangular $A\in M_{m,n}$ the same argument gives $t^{n}\chi_{AA^T}(t)=t^m\chi_{A^TA}(t)$ up to sizes: the non-zero eigenvalues of $AA^T$ and $A^TA$ coincide, with multiplicities, and the larger matrix has $|m-n|$ extra zero eigenvalues. These common values are the squares of the singular values of $A$.

Lesson. $XY$ and $YX$ are "almost conjugate": same characteristic polynomial even when neither $X$ nor $Y$ is invertible.