Use the recurrence D_n = n D_{n-1} + (-1)^n, n\geq 2 to find the exponential generating function for D_n.

Here is my solution: \sum_{n \geq 2} D_n \frac{x^n}{n!} = \sum_{n \geq 2} n D_{n-1} \frac{x^n}{n!} + \sum_{n \geq 2} (-1)^n \frac{x^n}{n!} \\<br />
\implies & G_D(x) = x \sum_{n \geq 2} D_{n-1} \frac{x^{n-1}}{(n-1)!} + (e^{-x} -e^{-1}-1)\\<br />
\implies & G_D(x) = x \sum_{n \geq 1} D_{n} \frac{x^{n}}{n!} + (e^{-x} -e^{-1}-1) \\<br />
\implies & G_D(x) = x (D_{1} \frac{x^{1}}{1!}+\sum_{n \geq 2} D_{n} \frac{x^{n}}{n!}) + (e^{-x} -e^{-1}-1) \\<br />
\implies & G_D(x) = x G_D(x) + (e^{-x} -e^{-1}-1) \\<br />
\implies & G_D(x) = \frac{e^{-x} -e^{-1}-1}{1-x} <br />

But when I looked at the solution, the -e^{-1}-1 was unnecessary.
This happened because I began the summation from 2.
I did this because (i) the problem stated n\geq 2, and (ii) D_{-1} wouldn't make sense.
Was this unnecessary?

Also, if the problem asked me to find "the generating function for n\geq 2", does that mean
G_D(x)=\sum_{n \geq 2} D_n \frac{x^n}{n!}
G_D(x)=\sum_{n \geq 0} D_n \frac{x^n}{n!}

Finally, if there are some advices to not to get confused with indices (some general rules for handling them or something), I would much appreciate them.