Smallest Number with 2^n Divisors/Mistake

Source Work

 * The Dictionary
 * $120$
 * $120$

Mistake

 * The smallest number having $2^n$ divisors is found by multiplying together the first $n$ numbers in this sequence: $2$, $3$, $4$, $5$, $7$, $9$, $11$, $13$, $16$, $17$, $19$, $\ldots$ which consists of all the primes and powers of primes.

Correction
The last clause should say:
 * ''... which consists of all the numbers of the form $p^{\paren {2^k} }$ where $p$ is prime and $k \ge 0$.