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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 $S$, $$e^Te^S=\lim_{n\to+\infty}\Big(I+\frac{T}{n}\Big)^n\Big(I+\frac{S}{n}\Big)^n=\lim_{n\to+\infty}\bigg(I+\frac{T+S+\frac{TS}{n}}{n}\bigg)^n=e^{S+T}\, .$$