16. Let the product of all distinct positive divisors of 2005 be $a$, and the product of all distinct positive divisors of $a$ be $b$. Then $b=$ $\qquad$
16.2 0059.

From $2005=5 \times 401$, we know that any positive divisor of 2005 is the product of one of $1, 5$ and one of $1, 401$, so $a=5^{2} \times 401^{2}$. Furthermore, any positive divisor of $a$ is the product of one of $1, 5, 5^{2}$ and one of $1, 401, 401^{2}$, hence
$$
b=\left(5 \times 5^{2}\right)^{3} \times\left(401 \times 401^{2}\right)^{3}=2005^{9} .
$$

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