To find the right most non zero digit When expanded 30! ends in 7 zeroes. Find the first non zero digit from right?
 A: First, consider the product of elements after dividing everything by all powers of 2 and 5.
$1\cdot 2 \cdot 3 \cdot \ldots \cdot 30$
becomes:
$1 \cdot 1 \cdot 3 \cdot 1 \cdot 1 \cdot 3 \cdot 7 \cdot 1 \cdot 9 \cdot 1 \cdot
11 \cdot 3 \cdot 13 \cdot 7 \cdot 3 \cdot 1 \cdot 17 \cdot 9 \cdot 19 \cdot 1 \cdot
21 \cdot 11 \cdot 23 \cdot 3 \cdot 1 \cdot 13 \cdot 27 \cdot 7 \cdot 29 \cdot 3$
Modulo 10, and ignoring ones, this becomes $3 \cdot 3 \cdot 7 \cdot 9 \cdot 3 \cdot 3 \cdot 7 \cdot 3 \cdot 7 \cdot 9 \cdot 9 \cdot 3 \cdot 3 \cdot 3 \cdot 7 \cdot 7 \cdot 9 \cdot 3$.
Writing 7 as $-3$ and 9 as $-1$, this becomes $3^9 \cdot (-3)^5 \cdot (-1)^4 = -(3^{14}) = -(9^7) = 1$.
Now, all powers of 2 and 5 are not missing from the product. 5 is clearly the bottleneck, so notice that power of 2 in this equals $15 + 7 + 3 + 1 = 26$. Out of these 7 are taken up in the zeroes at the end. So we are left with $2^{19}$. $2^{10} = 4 \bmod 10$, and $2^9 = 2 \bmod 10$. So $2^{19} = 8 \bmod 10$. As the previous product we found was 1, the digit you want is also 8.
A: 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 $n\equiv8\pmod{10}.$
A: The leading non-zero digit of $ n! $ is $ = [\frac{n}{5}]! \times {2}^{[ \frac{n}{5} ]} \times Rem[\frac{n}{5}]!$ where [] denotes the greatest integer function.
So the last non-zero digit of $ 30! = $ last non-zero digit of $[\frac{30}{5}]! \times {2}^{\frac{30}{5}} =$ last non zero digit of $ 6! \times {2}^{6} $
Now $ 6! = 720 $ which means the last non-zero digit of $6!$ is $2$.
So combining those we get the last non-zero digit of $ 30! = $ unit digit of $(2 \times 2^6 ) = $ unit digit of $(128) = 8 $.
Hence the answer is 8.
