Orthonormal basis construction for Hermetian matrix

Question

I was going thru wiki regarding this proof. I understood the following

Given a matrix $A$ of order $n$, its characteristic polynomial is of degree $n$ and so it will have $n$ roots(by fundamental theorem of algebra). Let the root be $\lambda_1$ and corresponding eigen vector be $\vec e_1$. Now consider $K = \text{Span} \{ \vec e_1\}^{\perp}$. Now again we have $n-1$ degree characteristic polynomial and so again (by above) we have an eigen value and corresponding eigen vector...

In this way we can construct orthonormal basis for $ A$(Hermetian or symmetric matrix of order $n$).

But in wiki "The spectral theorem holds also for symmetric maps on finite-dimensional real inner product spaces, but the existence of an eigenvector does not follow immediately from the fundamental theorem of algebra. To prove this, consider $A$ as a Hermitian matrix and use the fact that all eigenvalues of a Hermitian matrix are real." --> why the existence do not immediately follow from $n-1$ degree characteristic polynomial? It should right? kindly elaborate.

Wiki link - https://en.wikipedia.org/wiki/Spectral_theorem#:~:text=In%20mathematics%2C%20particularly%20linear%20algebra,diagonal%20matrix%20in%20some%20basis.&text=In%20more%20abstract%20language%2C%20the,about%20commutative%20C*-algebras.

in proof of this theorem - Theorem. If $A$ is Hermitian, there exists an orthonormal basis of V consisting of eigenvectors of $A$. Each eigenvalue is real....]