the number of zeros is $ \sum_{i = 1}^{\infty}{[\frac {10000}{5^i}]} = 2499$ so, mod 10: $ \frac {10000!}{10^{2499}}\equiv {(1*3*7*9)}^{1000}*2^{\sum_{i = 1}^{\infty}{[\frac {10000}{2^i}]} - 2499} \equiv 2^{7496} \equiv 6 \pmod {10}$ bye paolo

Unfortunately bboypa wolfram alpha is giving me an answer of 8, not 6. Your mistake is failing to account for what happens to the numbers that are divisible by 5 or 2 after you divide the 5's and 2's out.

Peter wrote: Counting from the right end, what is the $2500$th digit of $10000!$? We have to get $x\equiv \frac{10000!}{2^{2499}\cdot 5^{2499}}$ $mod 10$ Since $ord_{2}10000!>ord_{5}10000!=2499$,$x$ is even, namely $x\equiv 0,2,4,6,8$ $mod 10$. On the other hand, $\frac{10000!}{5^{2499}} \equiv (1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdot 8\cdot 9)^{100}\equiv 6^{1000}\equiv 6$ $mod 10$ Then $2^{2499}\cdot x\equiv 6$ $mod 10$ $\Leftrightarrow$ $x\equiv 2,7$ $mod 10$ Therefore $x\equiv 2$ $mod10$. The answer is $\boxed 2$. $\blacksquare$