Finite static fermion-mode trace and occupation identity

Let $M\ge 0$ be an integer, let $E_1,\dots ,E_M\in \mathbb{R}$, $\beta >0$, and $\mu \in \mathbb{R}$. Define the finite index set ${\mathcal{N}}_M=\{0,1{\}}^M$ and the finite-dimensional Hilbert space ${\mathcal{H}}_M={\mathbb{C}}^{{\mathcal{N}}_M}$ with orthonormal basis $\mathrm{\mid }{n}_1,\dots ,n_M⟩$. Define diagonal operators $H$ and $N$ on this basis by $H\mathrm{\mid }\mathbf{n}⟩=\left({\sum }_{j=1}^M{E}_j{n}_j\right)\mathrm{\mid }\mathbf{n}⟩,N\mathrm{\mid }\mathbf{n}⟩=\left({\sum }_{j=1}^M{n}_j\right)\mathrm{\mid }\mathbf{n}⟩\mathrm{.}$ For a matrix on ${\mathcal{H}}_M$, trace means the finite sum of its diagonal entries in this basis. Then $\mathrm{\Xi }=\mathrm{Tr}{e}^{-\beta (H-\mu N)}={\prod }_{j=1}^M(1+e^{-\beta (E_j-\mu )})\in (0,\mathrm{\infty })\mathrm{.}$ The normalized diagonal state with weights ${\mathrm{\Xi }}^{-1}{e}^{-\beta ({\sum }_j{E}_j{n}_j-\mu {\sum }_j{n}_j)}$ has, for each $1\le j\le M$, occupation mean $⟨n_j⟩=\frac{1}{e^{\beta (E_j-\mu )}+1}\mathrm{.}$ For $M=0$, the empty product and trace are $1$, and there are no occupation marginals. This is a finite-dimensional algebraic identity and uses no choice principle, curved-space spectrum, or gravitational field equation.