$\def\ad{\text{ad}}$
This follows immediately from the Baker-Campbell-Hausdoff formula:
For complex $z$ near $1$ we consider the function
$$f(z):= \frac{\log(z)}{z-1} = \sum_{n\geq0}\frac{(-1)^n}{n+1}(z-1)^n$$
Then for $X$, $Y$ near $0$ in $\mathfrak g$ we have $\exp X.\exp Y= \exp
C(X,Y)$, where
$$ C(X,Y) = Y + \int_0^1 f(e^{t. \ad X}.e^{ \ad Y}).X\,dt $$
$$
= X + Y + \sum_{n\geq1}\frac{(-1)^n}{n+1}\int_0^1\biggl(
\sum_{{k,\ell\geq0, k+\ell\geq1}} \frac{t^k}{k!\,\ell!}\;
( \ad X)^k( \ad Y)^\ell\biggr)^nX\;dt
$$
$$
= X + Y + \sum_{n\geq1}\frac{(-1)^n}{n+1}
\sum_{{k_1,\dots,k_n\geq0,
\ell_1,\dots,\ell_n\geq0,k_i+\ell_i\geq1}}
\frac{(\ad X)^{k_1}(\ad Y)^{\ell_1}\dots
(\ad X)^{k_n}(\ad Y)^{\ell_n}}
{(k_1+\dots+k_n+1)k_1!\dots k_n!\ell_1!\dots\ell_n!}X
$$
$$
= X + Y + \tfrac12[X,Y] +\tfrac1{12}([X,[X,Y]]-[Y,[Y,X]]) + \cdots .
$$
For a short proof of this formula see 4.29 of here.
This series has radius of convergence $\pi$ in each norm on the Lie algebra in which the bracket (as bilinear operator) is bounded by 1.
This even works for $C^2$-Lie groups, since $C^2$ suffices to get the Lie bracket at the tangent space of the identity.

allsmooth maps are real-an.). – BCnrd Dec 16 '10 at 1:03