jmc

By pairing up factors of 2 with factors of 5 to make factors of 10, the product may be written as $$\begin{array}{rl}{4}^5\cdot 5^{13}& =2^{10}\cdot 5^{13}\\ & =(2\cdot 5{)}^{10}\cdot 5^3\\ & =125\cdot 10^{10}.\end{array}$$ `125' followed by 10 zeros has $10+3=\boxed{13}$ digits.