Let A be an invertible matrix. Then A^-1 exists. Assume that A has a left inverse. Then A^-1 A = In. We need to show that A has a right inverse i.e. A A^-1 = In. Let B = A A^-1. We need to show that B = In. Multiply both sides of B = A A^-1 by A on the right. Then BA = (A A^-1) A = A (A^-1 A) = A In = A. Then multiply both sides on the right by A^-1. Then (BA)A^-1 = AA^-1. Then B(A A^-1) = A A^-1. Then BB = B. Then B^2 = B. Since A is invertible, det(A) =/= 0. Then det(A^-1) = 1/det(A) =/= 0, A^-1 is invertible. Then det (B) = det(A A^-1) = det(A) * det(A^-1) = det(A) * 1/det(A) = 1 =/= 0, so B is invertible. Then B^-1 exists and B has a left inverse. Then multiply both sides of B^2 = B by B^-1 on the left. Then B^-1 (B^2) = B^-1 B. Then (B^-1 B) B = B^-1 B. Then In B = In. Then B = In. Thus, A A^-1 = In. Thus, A has a right inverse.