Prove that if $A^{t}A$ is idempotent then $A^{t}=A^{+}$

Let $B=AA^TA$. Then $B^TB=A^TA=A^TB$. It follows that $\|B\|_F^2=\|A\|_F^2=\langle A,B\rangle$ and in turn, $\langle A,B\rangle=\|A\|_F\|B\|_F$. Hence $A=B=AA^TA$. Take transposes on both sides, we also get $A^T=A^TAA^T$.