jmc

Let's multiply the divisors of a positive integer, say $12$. The divisors of $12$ are $1,2,3,4,6,$ and $12$. The product of the divisors of 12 is $1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 12=(1\cdot 12)(2\cdot 6)(3\cdot 4)=12^3$. The factors may be regrouped in this way for any positive integer with an even number of divisors. We have found that if the number $d$ of divisors is even, then the product of the divisors of $n$ is $n^{d/2}$. Solving $n^6=n^{d/2}$, we find $d=12$.

Recall that we can determine the number of factors of $n$ by adding $1$ to each of the exponents in the prime factorization of $n$ and multiplying the results. We work backwards to find the smallest positive integer with $12$ factors. Twelve may be written as a product of integers greater than 1 in four ways: $12$, $2\cdot 6$, $3\cdot 4$, and $2\cdot 2\cdot 3$. The prime factorizations which give rise to these products have sets of exponents $\{11\}$, $\{5,1\}$, $\{3,2\}$, and $\{2,1,1\}$. In each case, we minimize $n$ by assigning the exponents in decreasing order to the primes $2,3,5,\dots$. Therefore, the smallest positive integer with 12 factors must be in the list $2^{11}=2048$, $2^5\cdot 3=96$, $2^3\cdot 3^2=72$, and $2^2\cdot 3\cdot 5=60$. The smallest of these is $\boxed{60}$.