= Solution
Use the real <vector space> of <Hermitian matrices> and its <Hilbert-Schmidt inner product> $\langle A,B\rangle=\operatorname{Tr}(AB)$. A Hermitian $N$-by-$N$ <matrix> has $N$ real diagonal entries and two real parameters for each off-diagonal pair, hence real dimension $N+2\binom N2=N^2$. Imposing zero <trace> removes one real dimension. Each supplied <generalized Pauli matrix> is Hermitian and traceless, and there are $N(N-1)$ off-diagonal matrices and $N-1$ diagonal matrices, totaling $N^2-1$.
Their assumed <orthonormality> makes them linearly independent. Since this equals the dimension of the traceless space, \b[they are an orthonormal basis of the traceless Hermitian matrices]. To extend the family, the required additional generator is $\sigma_0=I/\sqrt N$, which is implicit in the indexing of the question. It has unit <Hilbert-Schmidt norm> and is orthogonal to every traceless generator, so the extended family is an <orthonormal basis> of all Hermitian matrices.
For a <density operator> $\rho$, cyclicity of the <trace> gives
$$
s_k^*=\operatorname{Tr}((\rho\sigma_k)^\dagger)=\operatorname{Tr}(\sigma_k\rho)=\operatorname{Tr}(\rho\sigma_k)=s_k.
$$
Thus \b[$s\in\mathbb R^{N^2-1}$]. Since $\operatorname{Tr}\rho=1$, the resulting <Generalized Bloch representation> is
$$
\rho=\frac IN+\sum_{k=1}^{N^2-1}s_k\sigma_k.
$$
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