Differentiability at $a$ means that there is a linear map $L:V\to V_1$ withThis map is the total derivative $D{f}_a$. In the Frobenius norm, $|AB|\le |A||B|$, by the Cauchy-Schwarz inequality applied to matrix entries. Expanding with the order of all factors preserved givesThe remainder after the terms linear in $H$ has norm at most $3|A||H{|}^2+|H{|}^3=o(|H|)$. HenceFor $|C|<1$, the Neumann series $I-C+C^2-\cdots$ converges absolutely in the finite-dimensional space of matrices. Multiplying its partial sums by $I+C$ and passing to the limit shows that it is $(I+C{)}^{-1}$. Its remainder after $I-C$ has norm at most $|C{|}^2/(1-|C|)=o(|C|)$, proving the required expansion.

If $A$ is an invertible matrix and $|H|<1/|A^{-1}|$, then $A+H=A(I+A^{-1}H)$ is an invertible matrix. Thus the general linear group is an open subset of the matrix space. The preceding expansion also givesApply the chain rule to $g(X)=(X^{-1}{)}^3$, using the cubic total derivative at $A^{-1}$ and this total derivative of inversion:The three terms generally cannot be combined by commuting $A$ and $H$.

A topological space is compact if every open cover has a finite subcover. It is connected if it cannot be written as the union of two disjoint nonempty open subsets. It is Hausdorff if every two distinct points have disjoint open neighborhoods. These definitions apply to the space itself and, with the subspace topology, to its subsets.

Wilson's theorem states that an integer $n>1$ is prime if and only if $(n-1)!\equiv -1(\mathrm{mod}n)$. For a prime $p$, every nonzero residue has a multiplicative inverse. Pair each residue with its inverse; the only self-inverse ones satisfy $(a-1)(a+1)=0$ in ${\mathbb{F}}_p$ and hence are $1,-1$ (the same residue when $p=2$). The resulting product is $-1$. Conversely, if $(n-1)!\equiv -1(\mathrm{mod}n)$, no divisor $d$ of $n$ can satisfy $1<d<n$: it would divide both $(n-1)!$ and $n$, so could not divide $(n-1)!+1$. Hence $n$ is prime.

For a composite $n\ge 6$ which is not the square of a prime, choose a proper factorization $n=ab$ with $1<a<b<n$ if possible. Both factors occur in $(n-1)!$, proving divisibility by $n$. If the initially chosen factors coincide but their common value is composite, choose a different proper divisor to obtain unequal factors; equivalently, the only composite integer all of whose proper nontrivial divisors coincide is a prime square. For $n=p^2\ge 9$, the distinct factors $p$ and $2p$ lie below $p^2$ and their product is divisible by $p^2$. Thus $(n-1)!\equiv 0(\mathrm{mod}n)$ for every composite $n\ge 6$.

For a prime $n$, write $(n-1)!=kn-1$. Then ${\sin }^2((n-1)!\pi /n)={\sin }^2(\pi /n)$, so the summand is one. For a composite $n\ge 6$ it is zero. The only remaining composite index is four, whose summand equals ${\sin }^2(3\pi /2)/{\sin }^2(\pi /4)=2$. Therefore the sum is the prime-counting function plus two whenever $x\ge 4$; subtraction of two gives the claimed exact formula.