Can we assume that if for all values of , then for all ? Somehow I don't think this is valid, but let's run with it for a moment.
If this is true, then the whole point of the two-part exercise was to make the connection that for some where
Multiplying gives us . Now reduce the term and because the LHS is a whole number, the RHS must reduce fully. Since , 19 will still be a factor in this term after reducing, so it can be written for some positive integer k, which implies your hypothesis.
Not sure if this quasi-proof holds water, but it is the only connection I see between the two parts of the question.
I think I worked it out in my head now. One of the stipulations of the proof is that the function expressible by the series is infinitely differentiable in the interval of convergence. So defining , could not possibly be equal to zero within some interval and nonzero at a point outside this interval. Otherwise the nth derivative of h at would have a left hand limit of zero and a nonzero right hand limit for some .
*Not trying to argue with mathematical proof, I'm just trying to get my intuition to catch up to it. Lemma: No infinitely differentiable function can be constant on an interval no matter how small, unless it is constant everywhere.