jmc

For the product to be a perfect square, all the exponents need to be even. So we don't need to worry about factors that already have even exponents. We also don't need to worry about $9^9$ because $9$ is already a perfect square. The remaining factors are $3^3{5}^5{7}^7$.

To get even exponents in the product, we need at least one more $3$, at least one more $5$, and at least one more $7$. That would bring us up to $3^4{5}^6{7}^8$, and everything would be good. And indeed, $3\cdot 5\cdot 7=\boxed{105}$.