Invertible upper triangular matrices are diagonalizable.

Question

Is this 'result' true? I think it is so as we can choose the first column as pivot point and make all the entries of the first row zero (this can be done as all the diagonal entries are non zero to keep the determinant non zero). And then doing this repeatedly, we can make the whole matrix, a diagonal one. Also suggest an elegant proof if the 'result' is true.