Proving $\exp (T + S) = \exp (T) \exp (S) = \exp (S) \exp (T)$ for $T,S\in\mathcal{L}(E)$ 
Let $E$ be a Banach space and $T,S \in \mathcal L (E)$ be two commuting maps i.e., $TS = ST.$ Then, $$\exp (T + S) = \exp (T) \exp (S) = \exp (S) \exp (T).$$

I can able to prove it using the formula for exponential but the proof is very clumsy. Is there any easier way to prove the above equality? Any help in this regard will be appreciated.
Thanks in advance.
 A: Here's a different proof than the one linked (it also works in any Banach algebra). First, you can check that the power series $$e^{tS}=\sum\limits_{n=0}^\infty\frac{t^n}{n!}S^n$$ makes sense for all $t\in\mathbb{R}, S\in\mathcal{L}(E).$ Differentiating in $t$ gives $$\frac{d}{dt}e^{tS}=Se^{tS}=e^{tS}S.$$
Next, compute that $$\frac{d}{dt}(e^{t(S+T)}e^{-tT}e^{-tS})=e^{t(S+T)}Se^{-tT}e^{-tS}-e^{t(S+T)}e^{-tT}Se^{-tS}.$$ We claim this equals zero. Indeed, the fact that $ST=TS$ implies that $$e^{-tT}S=\sum\limits_{n=0}^\infty \frac{(-t)^n}{n!}T^nS=S\sum\limits_{n=0}^\infty \frac{(-t)^n}{n!}T^n=Se^{-tT},$$ from which it follows that $Se^{-tT}=e^{-tT}S$.
Hence, $$e^{t(S+T)}e^{-tT}e^{-tS}$$ is constant in $t$. Evaluating at $t=0$ gives that $$e^{t(S+T)}e^{-tT}e^{-tS}=I.$$ All that we must show is that the inverse of $e^{tS}$ is $e^{-tS},$ as the result will follow from inversion (multiply on the right by $e^{tS}$, then by $e^{tT}$). Indeed, differentiate $e^{(s+t)S}e^{-tS}$ in $t$. You'll get that it's zero, so it's constant in $t$, and evaluation at $t=0$ gives $$e^{(s+t)S}e^{-tS}=e^{sS}.$$ Evaluate this at $s=0$ to get the inversion property.
This shows that $$e^{t(S+T)}=e^{tS}e^{tT}.$$ Evaluating at $t=1$ gives the result that you wanted (well, the first equality; to get the other, simply follow the same argument with $e^{t(S+T)}e^{-tT}e^{-tS}$).
Alternatively, this results directly from e.g. the holomorphic functional calculus. This also works in any Banach algebra.
