= Solution
Use the <truncated tensor algebra> $T^{(N)}(\mathbb R^d)=\bigoplus_{k=0}^N(\mathbb R^d)^{\otimes k}$, with truncated concatenation product. The step-$N$ <path signature> is
$$
S_N(x)_{s,t}=1+\sum_{k=1}^N\int_{s<u_1<\cdots<u_k<t}dx_{u_1}\otimes\cdots\otimes dx_{u_k}.
$$
These <iterated integrals of a path> exist as <Riemann-Stieltjes integrals>, since a <Lipschitz continuous> path on a compact interval has <bounded variation>. The degree-zero coordinate is $1$, and the degree-one coordinate is $x_t-x_s$.
Let $S_t=S_N(x)_{0,t}$ and write $S_t^{(k)}$ for its degree-$k$ coordinate. Integrating first over all variables except the last gives
$$
S_t^{(k)}=\int_0^t S_u^{(k-1)}\otimes dx_u,\qquad S_t^{(0)}=1.
$$
Thus $dS_t^{(k)}=S_t^{(k-1)}\otimes dx_t$ for $1\leq k\leq N$. If $e_i$ is a standard basis vector, define the linear vector field $W_i(g)=g\otimes e_i$, with degrees above $N$ discarded. The component identities are precisely the controlled <ordinary differential equation>
$$
\boxed{dS_t=\sum_{i=1}^d W_i(S_t)\,dx_t^i=S_t\otimes dx_t,\qquad S_0=1.}
$$
For a Lipschitz driver this may also be written $\dot S_t=S_t\otimes\dot x_t$ almost everywhere. Recursive integration of its levels proves uniqueness and recovers the signature formula. The <Chen identity> follows by splitting each integration simplex at an intermediate time.
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