I want to prove that given any Hermitian operator, we can find an orthonormal eigen basis for it. It is obvious there are $n$ eigenvalues counting multiplicities, and it is easy to prove that any two distinct eigenvalues have orthogonal eigenvectors. But I run into trouble with algebraic multiplicities greater than one. If we assume that there are $n$ distinct eigenvectors for an eigenvalue with multiplicity of $n$, then we can again find orthogonal eigenvectors. But everywhere I look immediately assumes that there are $n$ distinct eigenvectors. Is it obvious that an $N \times N$ Hermitian matrix has $N$ eigenvectors?
Some proofs I've seen simply state that a Hermitian operator is diagonalizable, and thus must have n linearly independent eigenvectors with which the gram-schmidt can be applied for orthogonality. But when I look up proofs of diagonlizability, such as this one, it assumes we have $n$ orthogonal eigenvectors. So pretty much useless for me.
I may be missing something crucial here but I can't figure out how to prove there are $N$ linearly independent eigenvectors, which seems to amount to proving that the eigenspace of any eigenvalues with multiplicities of $n$, is $n$-dimensional. If someone knows of a way to prove this it would be greatly appreciated.