Using functional calculus to show that $exp(S)exp(T)=exp(T)exp(S)$

Question

Suppose $S$ and $T$ are two commuting normal operators in $B(H)$,how to use the continuous functional calculus to show that following conclusion?

$exp(S)exp(T)=exp(T)exp(S)=exp(S+T)$?

Can we derive the following formula:
$\exp (a) = \sum_{n=0}^\infty \frac{1}{n!} a^n,a\in B(H)$

Answer 1

What you do is to notice that $f(T)$ is in the C$^*$-algebra (von Neumann if $f$ is Borel) generated by $T$. Then $f(T)S=Sf(T)$. Now you repeat the argument, but with $f(T)$ and $S$ to get $f(T)g(S)=g(S)f(T)$.