Answer: 8

Solution:

Every factorization is $10^n = (2^x 5^y)(2^{n-x} 5^{n-y}).$ If a factor is divisible by both $2$ and $5,$ it is a multiple of $10$ and ends in the digit $0.$ So the only possible zero-free factorization is $10^n = 2^n \cdot 5^n,$ and we need the least $n$ for which $2^n$ or $5^n$ contains a digit $0.$

The powers $2^1, \ldots, 2^8$ are $2, 4, 8, 16, 32, 64, 128, 256$ — no zeros. The powers $5^1, \ldots, 5^7$ are $5, 25, 125, 625,$ $3125, 15625, 78125$ — no zeros — but $5^8 = 390625$ contains a $0.$

Hence every factorization of $10^8$ contains a digit $0,$ while $10^7 = 2^7 \cdot 5^7 = 128 \cdot 78125$ does not, so the answer is $8.$