Which one is the larger : $20!$ or $2^{60}$? Which one is the larger : $20!$ or $2^{60}$ ?

I am looking for an elegant way to solve this problem, other than my solution below. Also, solution other than using logarithm that uses the analogous inequalities below.
My solution:
Write $20!$ in prime factors and $2^{n}$:
$$ 20! = (2^{2} \cdot 5)(19)(2 \cdot 3^{2})(17)(2^{4})(3 \cdot 5)(2 \cdot 7)(13)(2^{2} \cdot 3)(11)(2 \cdot 5)(3^{2})(2^{3})(7)(2 \cdot 3) (5) (2^{2}) (3) (2) $$
$$ = 2^{18} (5)(19)(3^{2})(17)(3 \cdot 5)(7)(13)(3)(11)(5)(3^{2})(7)( 3) (5) (3) $$
so it is left to compare $2^{42}$ and $(5)(19)(3^{2})(17)(3 \cdot 5)(7)(13)(3)(11)(5)(3^{2})(7)( 3) (5) (3) $.
We write the prime factors nicely as:
$$ 3^{8}5^{4}7^{2}(11) (13)(17)(19) $$
Notice
$(3)(11) > 2^{5}$,
$(13)(5)>2^{6}$,
$(19)(7) >2^{7}$,
$17 > 2^{4}$, so we now focus on
$$3^{7}5^{3}7 = 2187(5^{3})7 > 2048(5^{3})7 = 2^{11}875 >2^{11}512 = 2^{20} $$
So we have that the prime factors is larger than $2^{42}$.

This answer extends the comment of zwim.
First, see
Stirling approximation For factorials.
I know that as $n$ goes to $\infty$, that the geometric mean of
$\{1,2,\cdots,n\}$ approaches $~\displaystyle \frac{n}{e}~$ from above.  Further, as $n$ increases, the ratio between $n$ and the geometric mean of $\{1,2,\cdots,n\}$ is strictly decreasing.
I also know, from involvement with a prior (similar) problem involving $(100)!$, that even with the number as large as $(100)$, the geometric mean of $\{1,2,\cdots,100\}$ is still greater than $(40)$.
This implies that since $(20) < (100)$, the geometric mean of $\{1,2,\cdots,20\}$ must be greater than $~(20 \times 0.4).~$
This surmise makes it game over, because $2^{60} = 8^{20}$.  Therefore, since the geometric mean of $\{1,2,\cdots,20\}$ is greater than $(8)$, you must have that $(20)! > 8^{20}.$
