At 0 isn't the function just 1, so the Taylor Polynomial would be 1...
Edit - I suppose you could keep differentiating and apply L'Hopital's to find the derivatives. But they still won't be zero.
And the question isn't badly worded, either. They meant they want the series centered at x=0, so the form will be , instead of
@OP: In this specific case, you start with the expansion for e^x:
(notice how i simply adjusted the index to ommit the constant term)
then divide by x:
Can you work from there?
For the remainder, unfortunately I haven't worked too much with that so I can't confidently provide an answer, but this might help:
Taylor's theorem - Wikipedia, the free encyclopedia