Prove $\exp(x)\exp(y)=\exp(y)\exp(x)$ if $[x,y]=0$. How to prove the following statement

Prove that $\exp(x)\exp(y)=\exp(y)\exp(x)$ if $x,y\in \mathfrak{g}$ and $[x,y]=0$.

by using the following results:


*

*$\mathrm{ad} (x)\cdot y=[x,y]$;

*$\mathrm{Ad} (\exp(x))=\exp(\mathrm{ad}x)$.


where $\mathrm{Ad}:G\rightarrow \mathrm{Gl}(\mathfrak{g})$ and $\mathrm{ad}:=\mathrm{Ad}_{\ast}:~\mathfrak{g}\rightarrow \mathfrak{gl}(\mathfrak{g})$.
(Without using the Campbell-Hausdorff formula...)

Attempt: 
If $[x,y]=0$, then $\mathrm{ad}(x)\mathrm{ad}(y)=\mathrm{ad}(y)\mathrm{ad}(x)$.
Also, we have $\mathrm{Ad}(\exp(xy))=1$. But I cannot move forward.
 A: I use $e^x = \exp(x)$. Note that 
$$e^x e^y = e^y e^x$$
is the same as $c(e^y)(e^x) = e^x$, where $c(g) : G \to G$ is the conjugation $c(g) h = g^{-1} hg$. Note that by definition $Ad(g) = c(g)_* :\mathfrak g \to \mathfrak g$. 
Note from (1) and (2)
\begin{equation}
\begin{split}
Ad(e^y) (x) &= \operatorname{Exp}(ad (y)) (x)\\ 
&= x + ad(y)x + \frac{1}{2} ad(y)^2 x + \frac{1}{3!} ad(y)^3 x + \cdots \\
&= x
\end{split}
\end{equation}
since $ad(y) x = 0$. 
Write $g = e^y$. From $Ad(g) x = x$ we want to integrate to say that $g^{-1} e^x g = e^x$. Consider 
$$\gamma(t) = g^{-1} e^{tx} g,$$
Note that by definition of $Ad$, $\gamma'(0) = Ad(g)x = x$. 
Also $\gamma(0) = e_G$ and $\gamma(t_1+ t_2) = \gamma(t_1)\gamma(t_2)$. In particular, 
$$\gamma'(t) = \gamma(t)_* \gamma'(0) = \gamma(t)_* x.$$ 
Thus $\gamma(t)$ and $e^{tx}$ are both the integral curve of the vector fields $X (g)= g_* x$ (just the left invariant vector fields defined by $x\in \mathfrak g$) which start at the identity $e_G$. Thus 
$$ \gamma(t) = g^{-1} e^{tx} g = e^{tx}$$
for all $t$ and in particular $g, e^x$ commutes. 
