Find natural number n such that (5n)! has 2018 trailing 0's?

Question

What is the least positive integer n that can be placed in the following expression:

n!(n+1)!(2n+1)! - 1

and yields a number ending with thirty digits of 9's.

Answer 1

1914

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Lets find out how zero's formed:
5!- 1 zero,
10!- 2 zero's
90!- 2*9=18 zero's
95!- 19 zero's
100!- 19+2=21 zero's
900!- 21*9=189 zero's
995!- 189+19=208 zeros
1000!- 208+3=211 zero's
9000!- 211*9=1899 zero's
Now we need 2018-1899=119 which we can get from 5*21+7*2 which is 570, so total is 9570! will give us 2018 zero's
5n = 9570 => n = 1914