I haven't checked your formulae, it looks long
However, note that the power series of the exponential starts at n=0, and the power series of the logarithm stars at n=1.
This should solve the problem of dividing by 0.
The MacLaurin series is a power series. The expansion of a function into a series is unique. So expanding a function as a power series is equivalent to getting the MacLaurin series. (I don't know if I'm clear here ? :s)
To simplify calculations, you can use the Cauchy product for series. But again, be careful that the series should start at the same point