We know that $A^T A$ is a symmetric matrix but is it always non-negative definite? (I fear not but it's worth asking!)
Now let's say $S$ is a skew matrix, hence we know that it's not diagonalizable in $\mathbb R^n$ but we might want to get the singular value decomposition if $S^T S$ is non negative definite. I have a feeling that it's always true... is there any theorem stating that?
If $A^TA$ is positive semidefinite, which I believe to be an equivalent form of your statement, then for all $x$, $x^TA^Tx \geq 0$. Now consider the following$$x^TA^TAx = (Ax)^T(Ax) = Ax \cdot Ax = ||Ax||_2^2$$ We know that all norms are non-negative values so it must be true that $||Ax||_2 \geq 0$. Thus, for all $x$, $x^TA^TAx \geq 0$ and $A^TA$ must be positive semidefinite.
The step that might be confusing, at least it was for me at first, is $(Ax)^T(Ax) = Ax \cdot Ax$ but remember that the matrix-vector product $Ax$ yields a vector.