Srikant asks for a "more elegant approach." Perhaps the following will respond to this challenge.
Let the argument of the exponential be $f(x)$ (so its power series coefficients are the gammas) and let the right hand side be $g(x)$ (with deltas as its coefficients), so that by definition
$$g(x) = \exp(f(x)).$$
Differentiating both sides and replacing $\exp(f)$ with $g$ yields
$$g' = f' * g.$$
Writing this out as power series gives the desired result: the delta comes from $g'$ while the convolution of the gammas and deltas comes from $f' * g$.
You don't have to worry about convergence (and the whole machinery of Taylor series), by the way: all these calculations can be performed in the ring of formal power series.