Matrix equation $(Ax - y)^T \cdot A = 0_n^T$ solve for $x$ I have the following matrix equation
$$(A x - y)^T \cdot A = 0_n^T$$
Whereby $x \in \mathbb{R}^n$, $y \in \mathbb{R}^d$, $A \in \mathbb{R}^{d \times n}$. A is not invertible, but $A^TA$ is invertible and the task is to solve for $x$. 
$0_n$ is a $n$-dim vector of 0s, $(0,0,\dots,0)^T$ and $^T$ is the transpose of a matrix.
In order to solve it I transformed the equations in the following way:
\begin{align*}
(Ax - y)^T \cdot A &= 0_n^T\\
(x^TA^T-y^T) \cdot A &= 0_n^T\\
x^TA^T \cdot A - y^T A &= 0_n^T\\
x^TA^T \cdot A &= y^T A\\
x^T &= (y^T \cdot A) \cdot (A^TA)^{-1}\\
x   &= ((y^T \cdot A) \cdot (A^TA)^{-1})^T\\
x   &= A^T \cdot y \cdot ((A^TA)^{-1})^T
\end{align*}
Is this legal or have I missed something?

Up to the sixth line of displayed equations, you are fine, thus $x=(y^TAB^{-1})^T$ with $B=A^TA$ (and $B$ is assumed to be invertible hence $B^{-1}$ exists), that is, $x=(B^{-1})^TA^Ty$ (do not forget that $(MN)^T=N^TM^T$, if only for dimension reasons). 
Now, $(B^{-1})^T=(B^T)^{-1}$ (this always holds) and $B^T=(A^TA)^T=A^T(A^T)^T=A^TA=B$ hence 
$$
x=B^{-1}A^Ty,\qquad B=A^TA.
$$
Dimensional analysis: The matrix $A$ has dimension $d\times n$ hence $A^T$ has dimension $n\times d$ and $B=A^TA$ has dimension $n\times$ d $\times$ d $ \times n=n\times n$, as well as $B^{-1}$. As a column vector, $y$ has dimension $d\times1$ hence $B^{-1}A^Ty$ has dimension $n\times$ n $\times$ n $\times$ d $\times$ d $\times 1=n\times 1$. All is well.
