Find a value of $n$ that has exactly 32 divisors I know that I could simply multiply the first $32$ primes together but is there some other way to ascertain the answer to this number theory problem?

Multiplying the first $32$ primes together would not work. Even multiplying the first three primes together gives you a number: $30$, with $8$ divisors: $\\{1, 2, 3, 5, 6, 10, 15, 30\\}$.

The number of divisors of a number is a multiplicative function, and since each prime has $2$ divisors, the product of $n$ distinct primes has $2^n$ divisors. That fact points to an answer to your original question.