I have to proove that $A \; n \times n$ is diagonalizable iff its eigenvectors form a basis of $\mathbb{C}^n$.
$(\to)$ If $A$ is diagonalizable then $A = SBS^{-1}$. Then $$A = SBS^{-1} \leftrightarrow AS=SD \leftrightarrow ASe_i=SDe_i = d_i (Se_i)$$ The set $V = \{Se_1, ..., Se_n\}$ is the set of column of $S$. Each column of $S$ is an eigenvector of $A$. Also, $S$ is invertible. It means $rkS=n$, so the set $V$ form a basis of $\mathbb{C}^n$.
$(\leftarrow)$ ?
My question is: how can I prove the second part of the question?
If we take a basis then the vectors of the basis are linearly independent, so if we take the vectors as column vectors of a matrix $A$ then the matrix is of full rank. We find the corresponding characteristics equation and all roots of the equation are its eigen-values of the matrix. Also all roots belong, as $\Bbb C$ is algebraically closed.
Once you got the eigen-values you can easily get the corresponding eigen-vectors and create the corresponding matrix $P$. $D$ the diagonal matrix consists the eigen-values in its diagonal entry. Thus $$A = P^{-1}DP$$
Hence $A$ is diagonalizable.