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For any prime $p,$ $(p-1)! \not\equiv 0 \pmod{p},$ so $\binom{p^9 - 1}{p-1} = \frac{(p^9-1)(p^9-2)\cdots(p^9-(p-1))}{(p-1)!} \equiv \frac{(p-1)!}{(p-1)!} \equiv 1 \pmod{p}$

It follows that $\begin{aligned} \frac{1}{p^9}\binom{p^9}{p} &= \frac{1}{p^9} \cdot \left(\frac{p^9}{p}\cdot \binom{p^9-1}{p-1}\right) \\ &= \frac{1}{p} \cdot \binom{p^9-1}{p-1} \\ &= \left\lfloor\frac{1}{p} \binom{p^9-1}{p-1}\right\rfloor + \frac{1}{p} \\ &= \left\lfloor\frac{1}{p} \binom{p^9-1}{p-1}\right\rfloor + \frac{p^8}{p^9} \end{aligned}$ so that $\binom{p^9}{p} \equiv p^8 \pmod{p^9}.$

Using $p=2999$ yields $\binom{2999^9}{2999} \equiv 2999^8 \pmod{2999^9}$ $\implies N = 2999^8 \equiv (-1)^8 = 1 \pmod{1000}$ showing that the last three digits of $N$ are $\boxed{001}$