To find the right most non zero digit When expanded 30! ends in 7 zeroes. Find the first non zero digit from right?

We desire the last digit of $n=\dfrac{30!}{10^7}$.
Note that $n\equiv0\pmod2$. Now we just need to find $n\pmod5$. By Wilson's theorem, or direct computation, we know $1(2)(3)(4)\equiv-1\pmod5$. This gives us 
$$\dfrac{30!}{5^7}\equiv (-1)\frac{5}{5}(-1)\frac{10}{5}\cdots\frac{25}{5^2}(-1)\frac{30}{5}\equiv(-1)^61(2)(3)(4)(6)\equiv(-1)^7\equiv -1\equiv 4\pmod5$$
However, we also have to take into account the factor of $2^7\equiv2(-1)^3\equiv-2\equiv3\pmod5$. Since the integers modulo $5$ are a field, we can divide $4$ by $3$ in this field to reveal $n\equiv 3\pmod5$
Since $n\equiv0\pmod2$ and $n\equiv3\pmod5$ we get $n\equiv8\pmod{10}.$
