Commutator $[\hat{A},\exp(\hat{A})]= 0$ In Equation (4) of this Physics.SE post, Qmechanic wrote, 
$$\tag{4}\frac{d}{dt}e^{t\hat{A}}~=~\hat{A}e^{t\hat{A}}~=~e^{t\hat{A}}\hat{A}.$$
How does one get this equation?
 A: All operators commute with themselves (obviously) and, for the same reason, any power of themselves. 
Therefore, continuing this line of argument, if a function of an operator has a uniformly convergent Taylor series, the function of the operator must commute with the operator itself.
Going even further: a function of an operator $f(t\,\hat{A})$ under less restrictive conditions to the above can be approximated to any degree of accuracy by a polynomial in $t\,\hat{A}$. Otherwise put: the set of polynomials in $t\,\hat{A}$ is dense in many suitably defined classes of functions of operators. So one can use this denseness, together with the continuity of operator composition, to show that $f(t\,\hat{A})$ and $\hat{A}$ commute for such function classes too.
So what this tells you is that in any expression involving only functions of one operator alone together with arithmetic operations of these functions, you can manipulate the expression exactly as though the one operator along were a scalar or an object in a commutative algebra.
A: The exponential of an operator is defined though a power series:
$$e^{t\hat{A}}=\sum^{\infty}_{n=0}\frac{(t\hat{A})^n}{n!}$$
Since the operator $\hat{A}$ commutes with itself, it commutes with the exponential:
$$\hat{A}e^{t\hat{A}}=\hat{A}\sum^{\infty}_{n=0}\frac{(t\hat{A})^n}{n!}=\sum^{\infty}_{n=0}\hat{A}\frac{(t\hat{A})^n}{n!}=\sum^{\infty}_{n=0}\frac{t^n\hat{A}^{n+1}}{n!}=\sum^{\infty}_{n=0}\frac{(t\hat{A})^n}{n!}\hat{A}=e^{t\hat{A}}\hat{A}$$
See Rod Vance's answer for the bigger picture.
