Core idea

For s<ts<t, the ordered integration simplex is

{(t1,…,tn):s<tn<⋯<t1<t}.\{(t_1,\ldots,t_n):s<t_n<\cdots<t_1<t\}.

Its volume is (t−s)n/n!(t-s)^n/n!. Up to the measure-zero sets where coordinates coincide, the cube (s,t)n(s,t)^n splits into n!n! congruent regions, one per coordinate ordering; dividing its volume proves the formula.

Iterated-integral bound

If ∥A(τ)∥≤M\|A(\tau)\|\le M, submultiplicativity of the operator norm gives

∥∫s<tn<⋯<t1<tA(t1)⋯A(tn) dtn⋯dt1∥≤(M(t−s))nn!.\left\|\int_{s<t_n<\cdots<t_1<t}A(t_1)\cdots A(t_n)\,dt_n\cdots dt_1\right\| \le\frac{(M(t-s))^n}{n!}.

The factorial records time ordering and makes the resulting series converge on every finite interval.