Smallest Number with 2^n Divisors/Mistake

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Source Work

1986: David Wells: Curious and Interesting Numbers:

The Dictionary
$120$


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 $k \ge 0$.


Sources