Prove that if A is a square matrix with linearly indep columns, then $A^+$ = $A^{-1}$

Question

I have figured out a possible solution and I only need someone to tell me if I'm correct. The solution is the following:

If a matrix has linearly independent columns, then $A^+$ = $(A^TA)^{-1}A^T$

Therefore, by operating: = $A^{-1}(A^T)^{-1}A^T$ = $A^{-1}$ since $(A^T)^{-1}A^T = I$ So $A^+ = A^{-1}$

Am I right with this proof?

Thank you

Answer 1

If you're allowed to assume that $A^+ = (A^TA)^{-1}A^T$, then your proof is indeed correct.