Taylor's theorem implies existence of n+1 order derivative?

Question

From my Calculus textbook:

Usually the text is careful to make sure the description above a formula establishes all the necessary preconditions to exist for the formula to be true. I noticed here though that $f^ {(n+1)} $ is referred to, without any mention of a requirement that it exist (f being order n differentiable by itself doesn't imply n+1 differentiability). Does the continuity of all the earlier derivatives imply it or is there an unstated assumption?

Answer 1

The statement says

$f^{(n)}$ is differentiable

so that next derivative exists