[Number Theory]
Problem: Compute the least integer $n>1$ such that the product of all positive divisors of $n$ equals $n^{4}$.

Solution: Note that every factor pair $d$ and $\frac{n}{d}$ have product $n$. For the product of all such divisor pairs to equal $n^{4}$, there must be exactly 4 divisor pairs, or 8 positive integer divisors. A number has 8 positive integer divisors if it is of the form $a^{3} b^{1}$ or $a^{7}$ where $a$ and $b$ are distinct primes. The prime factorization $a^{3} b^{1}(a \neq b)$ provides a set of divisors each of which has 4 options for using $a\left(a^{0}, a^{1}, a^{2}, a^{3}\right)$ and an independent 2 options for using $b\left(b^{0}, b^{1}\right)$. Using the least values $(a, b)=(2,3), a^{3} b^{1}=24$. If instead the prime factorization is $a^{7}$ (having divisors $a^{0}, a^{1}, a^{2}, \ldots, a^{7}$ ), the least answer would be $2^{7}=128$. Thus the answer is 24 .

Final answer: 24