TITLE: exp(S) exp(T) = exp(S+T) for commuting operators
QUESTION [11 upvotes]: The standard way to prove the exponential law for two bounded commuting operators $S, T$
$$
\exp(S)\exp(T) = \exp(S+T)
$$
is to pass by the binomial formula and the power series of $\exp(.)$. I wonder whether one can prove it alternatively by a (Dunford-Riesz) functional calculus argument i.e. a resolvent-based proof, either directly, via
$$
\exp(S)\exp(T) = \tfrac{1}{(2\pi i)^2 } \int_{|z|=r} \int_{|\lambda|=R} e^{\lambda+z} R(\lambda, S) R(z, T)\,d\lambda dz
$$
for some $r> \|T\|, R > \max(r, \|S\|)$ or maybe by an operator matrix approach (for example the functional calculus of $A= \left( \begin{array}{rr} S & I \\ 0 & T \end{array}\right)$).

REPLY [4 votes]: Also: for bounded operators,  $(I+T/n)^n$ converges to $e^T$ uniformly on bounded sets (however, uniform convergence on compact sets will suffice). So if $T$ commutes with $S$,  $$e^Te^S=\lim_{n\to+\infty}\Big(I+\frac{T}{n}\Big)^n\Big(I+\frac{S}{n}\Big)^n=\lim_{n\to+\infty}\bigg(I+\frac{T+S+\frac{TS}{n}}{n}\bigg)^n=e^{S+T}\, .$$