How does the spectral theorem imply that a normal matrices have an orthonormal basis of eigenvectors?

Question

Spectral Theorem:

Let $A \in \mathbb{C}^{n \times n}$

Then: $A$ is normal $\iff$ there exists a unitary matrix $U$ and a diagonal matrix $\Lambda$ s.t $$A=U\Lambda U ^*$$

I know that the diagonal matrix has A's eigenvalues as its diagonal entries.

A proof through which I'm working state that from the spectral theorem we have that normal matrices have an orthonormal basis of eigenvectors.

But how exactly does the spectral theorem establish this?

My approach at the moment.

(trying to show that an arbitrary column of U is an eigenvector of A, using the fact that U is unitary(orthogonal if working in reals) $A=U\Lambda U^* \iff AU= \Lambda U$

Let $U= [u_1 \cdots u_n]$ be U partitioned in terms of its columns.

Then equating the jth columns:

$$Au_j=\Lambda u_j$$.

This is where i get stuck at the moment!

Answer 1

$A=U\Lambda U^* \iff AU= \Lambda U$

This is incorrect : as $U^*U = I$, we have : $$A = U\Lambda U^* \iff AU = U\Lambda$$

Then, writing $\Lambda = \text{diag}(\lambda_1,\ldots, \lambda_n)$ and equating the columns, we have : $$Au_j = \lambda_j u_j$$