Clever way to show a property of Lie transformation Given the following Lie transformation:
$$
\exp(\lbrace H,  \cdot \rbrace):=\sum_{n=0}^{\infty} \frac{(\lbrace H,  \cdot \rbrace)^n}{n!}
$$
and apply it to a Poisson Bracket $\lbrace g_1,  g_2 \rbrace$.
I would like to show that
$$
\exp(\lbrace H, \lbrace g_1,  g_2 \rbrace\rbrace)=\sum_{n=0}^{\infty} \frac{(\lbrace H,  \lbrace g_1,  g_2 \rbrace \rbrace)^n}{n!}=\dots=\lbrace \sum_{n=0}^{\infty} \frac{(\lbrace H,  g_1\rbrace)^n}{n!}, \sum_{n=0}^{\infty} \frac{(\lbrace H,  g_2\rbrace)^n}{n!}\rbrace=\lbrace\exp(\lbrace H, g_1\rbrace),\exp(\lbrace H, g_2\rbrace)\rbrace
$$
I put some dots in the Passage I am not able to do...(I tried expressing the power using the binomial theorem)
I am sure this property holds, since I found this here (page 25): http://www.aps.anl.gov/Science/Publications/lsnotes/content/files/APS_1418211.pdf
but I have difficulties in proving this in an elegant way.
Could someone give me a good reference?
This is the way I tried:
$$\exp(\lbrace H, \lbrace g_1,  g_2 \rbrace\rbrace)=\sum_{n=0}^{\infty} \frac{(\lbrace H,  \lbrace g_1,  g_2 \rbrace \rbrace)^n}{n!}=
\sum_{n=0}^{\infty} \frac{(  \{\{H,g_1\},g_2\}+\{g_1,\{H,g_2\}\}   )^n}{n!}=
\sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^n {n \choose k}(\{\{H,g_1\},g_2\})^{n-k}(\{g_1,\{H,g_2\}\})^k = 
$$ 
$$\sum_{n=0}^{\infty} \sum_{k=0}^n \frac{1}{(n-k)!}(\{\{H,g_1\},g_2\})^{n-k}\frac{1}{k!}(\{g_1,\{H,g_2\}\})^k $$
 A: Although OP strictly speaking writes something different in the question formulation (v4), it seems OP essentially wants to prove the following Lemma. 

Lemma. Given a (possibly infinite-dimensional$^1$) Lie algebra $L$ over a field $\mathbb{F}\supseteq\mathbb{R}$ with Lie bracket $[\cdot,\cdot]:L\times L \to L$ and Lie algebra derivation$^2$ $D:L\to L$, which satisfies Leibniz rule
  $$ \forall x,y\in L:~~ D[x,y]~=~ [Dx,y]+[x,Dy].\tag{1}$$
  Assume that the exponential map 
  $$e^D:= {\bf 1}_L+ D + \frac{1}{2}D^2+ \ldots ~:~L\to L\tag{2}$$ 
  is pointwise convergent.
  Then 
  $$ \forall x,y\in L:~~ e^D[x,y]~=~[e^Dx,e^Dy].\tag{3}$$

Proof of eq. (3): The identity (3) follows by setting $t=1$ in the following identity:
$$ \forall t\in [0,1]\forall x,y\in L:~~ e^{tD}[x,y]~=~[e^{tD}x,e^{tD}y].\tag{4}$$
To prove eq. (4), first note that eq. (4) is trivially true for $t=0$. Next show that the lhs. and the rhs. of eq. (4) independently satisfy the same first-order ODE
$$ \frac{d}{dt}(\ldots)~=~D(\ldots). \tag{5}$$
Hence they must be equal.
--
$^1$ Note that the Poisson algebra $(C^{\infty}(M);\{\cdot,\cdot\}_{PB})$ of smooth functions over a finite-dimensional phase space $M$ is an infinite-dimensional Lie algebra.
$^2$ In OP's case the derivation $D=[z,\cdot]$ is an inner derivation.
