The last six digits of the number $30001^{18} $ 

You want to compute the last six digits of $(3 \cdot 10^4+1)^{18}$. Using the binomial formula, you have \[ (3 \cdot 10^4+1)^{18} = \sum_{k=0}^{18} \binom{18}{k} (3 \cdot 10^4)^k \] but now for $k \geq 2$ all the terms have at least $8$ zeros, so if you're interested in the last $6$ digits it's enough to look at the terms with $k=0$ and $k=1$.

In the end, we have \[ (3 \cdot 10^4+1)^{18} \equiv \binom{18}{0} \cdot 1 + \binom{18}{1} (3 \cdot 10^4) \equiv 18 \cdot 3 \cdot 10^4 + 1 \equiv 540001 \pmod{10^6} \] so the last six digits are $540001$.