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

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.
