jmc

We want to pair together the twos and the fives to make tens: $$\begin{array}{rl}{2}^3\cdot 3^1\cdot 4^3\cdot 5^8& =2^3\cdot 3\cdot 2^6\cdot 5^8\\ & =2^9\cdot 3\cdot 5^8\\ & =10^8\cdot 2\cdot 3\\ & =6\cdot 10^8\end{array}$$Therefore, we have $6$ with $8$ zeros at behind it, giving us $1+8=\boxed{9}$ digits.