Showing $(A^T A)^2 = A^T A$ , where matrix A has orthornomal rows

Question

Let $n > k$ be positive integers and let $A \in M_{k×n}(\mathbb{R})$ satisfy the condition that its set of rows is orthonormal with respect to the dot product. Show that $(A^T A)^2 = A^T A$.

I'm not sure if my proof is correct or there is any better proof for this problem. I appreciate any help.

Let $0\neq v\in\mathbb{R}^n$.

Since the rows of $A$ are orthonormal, then we get $AA^T=I_{k\times k}$, hence $\langle A^TAv, A^TAv\rangle=\langle v,(A^TA)^* A^TAv\rangle=\langle v,A^TAA^TAv\rangle=\langle v, A^TAv \rangle $. Note that $(A^T)^*=A.$ As, $v\in \mathbb{R}^n$ is arbitrary, we get $(A^T A)^2 = A^T A$.

Answer 1

HINT: $(A^T A)(A^T A) = A^T (A A^T) A$.

Show that the fact that $A$ has orthonormal rows means that the middle matrix is the identity. That's almost immediate from the definition of the matrix product and the word "orthonormal".

Answer 2

The set of rows of $A$ being orthonormal, means the set of columns of $A^\top$ is orthonormal as well. So we have $$ A^T = (b_1, \dotsc, b_k) \in \mathbb{R}^{n\times k} $$ with $$ b_i \cdot b_j = b_i^T b_j = \delta_{ij} $$ which means $$ A A^T = I $$ so we have $$ (A^T A)^2 = (A^T A)(A^T A) = A^T(A A^T) A = A^T I A = A^T A $$