NPN Circuit simulation doesn't agree with book
Your question: "What if Ic is sufficiently large to produce a voltage drop (on the resistor) bigger than Vcc?"

This is impossible. Using the given DC supply values (+- 9V) a voltage drop larger than Vcc would make the emitter potential positiv. In this case, the B-E pn junction is not positively biased and the transistr would be off. Hence, no current. That means: Your primary assumption (large Ic) is obsolete.

You should (a) know the "secrets of negative feedback (caused by R2) and (b) realize that such a negative DC feedback allows an automatic Ic adjustment because there are two equations (rules) which must befulfilled:  

(1) Ohms law for R2 and (2) Shockleys exponenrtial equation describing the Voltage-current relation across the B-E junctin.

Finally, for Vcc>0 one of the preconditions for using the BJT as an amplifier is met (C-B junction reverse-biased). The second condition (B-E junction forward-biased) is also met because the base potential (zero) is always larger than the negative collector potential.