The answer is 8.

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Note that $10^n = 2^n\times 5^n$ . Thus, when factored, the two factors will both be of the form $2^a5^b$ . If both $a>0$ and $b>0$ then $2^a5^b$ will be divisible by $10$ and thus contain a zero. So we only need to consider factorizations of the form $10^n = (2^n) \times (5^n)$ . $\begin{array}{c|c|c} n & 2^n & 5^n \\ \hline 1 & 2 & 5 \\ 2 & 4 & 25 \\ 3 & 8 & 125 \\ 4 & 16 & 625 \\ 5 & 32 & 3125 \\ 6 & 64 & 15625 \\ 7 & 128 & 78125 \\ 8 & 256 & 390625 \end{array}$

Thus $n=\boxed{8}$ is the smallest $n$ for which every factorization of $10^n$ contains a number with a 0.