Answer     8

Let's first factorize: $$30! = 2^{26}\times 3^{14} \times 5^7 \times 7^4 \times 11^2 \times 13^2 \times 17\times 19\times 23\times 29$$

Hence, $30!$ will end with $7$ zeros. Removing these zeros will get $$N=\frac{30!}{10^7}=2^{21}\times 3^{14} \times 7^4 \times 11^2 \times 13^2 \times 17\times 19\times 23\times 29$$

The desired result is the last digit of $N$: $$\begin{align*} N & \equiv 2^{21}\times 3^{14}\times 7^4\times 1^2\times 3^2\times 7\times 9\times 3 \\ &\equiv 2^{21} \times 3^{19}\times 7^5 \\ &\equiv\boxed{8} \pmod{10} \end{align*}$$