No. Smooth merely means infinitely differentiable, whereas analytic means that it has to be locally equal to its power series!
(But if we're talking complex differentiation, then being once differentiable is sufficient for meing analytic. Complex numbers are magical!)
The classic counterexample is the function:
[tex]
f(x) :=<br />
\begin{cases}<br />
e^{-1/x^2} & x \neq 0 \\<br />
0 & x = 0<br />
\end{cases}[/tex]
which is infinitely differentiable at x=0: in fact, we have that [itex]f^{(n)}(0) = 0[/itex] for all of its derivatives!
So, this function is clearly not equal to a power series on any neighborhood of zero, and thus is not analytic there.