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 $2^{42}$.
 A: $$20!=2^{18}\cdot3^8\cdot 5^4\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19$$
$$19\cdot 17\cdot 13=4199>2^{12}, \\3\cdot 11>2^5$$
So it suffices to show:
$$3^7\cdot 5^4\cdot 7^2>2^{25}.$$
Now, $5\cdot 7>2^5,$ and $3^2>2^3,$ so it suffices to showing:
$$75=3\cdot 5^2>2^{6}=64$$
This shows: $$\frac{20!}{2^{60}}=\left(1+\frac{103}{4096}\right)
\left(1+\frac1{32}\right)\left(1+\frac3{32}\right)^2\left(1+\frac18\right)^3\left(1+\frac{11}{64}\right)$$
A: 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 $(20)! > 8^{20}.$
A: Here is a way to go for the solution using only elementary computations.
(And building partial products that get bigger than and closer to powers of two. I wanted to use first very close approximations like $3\cdot 18\cdot 19=1026>1024$, but there is no need to be so economical at the beginning, and very generous at the end...)
$$
\begin{aligned}
3\cdot 12 &= 36 \\
&> 32 =2 ^5\ ,\\
5\cdot 13 &= 65 \\
&> 64 =2 ^6\ ,\\
6\cdot 7\cdot 9\cdot 11 &= 6\cdot 11\cdot 7\cdot 9 
=66\cdot 63=(64+2)(64-1)=64^2 +64-2
\\
&> 64^2 = (2^8)^2=2^{12}\ ,
\\
15\cdot 17\cdot 14\cdot 19
&=
(256-1)(16-2)(16+3)=
(256-1)(256+16-6)
\\
&>(256-1)(256+2)=256^2 + 256-2\\
&>256^2=(2^8)^2=2^{16}\ ,
\\
10\cdot 18\cdot 20 &= 3600
\\
&>2048=2^{11}\ ,
\\[3mm]
&\text{Putting all together:}
\\[3mm]
20!
&=2\cdot 4\cdot 8\cdot 16
\cdot(3\cdot 12)
\cdot(5\cdot 13)
\cdot(6\cdot 7\cdot 9\cdot 11)
\\
&\qquad\qquad\qquad
\cdot(14\cdot 15\cdot 17\cdot 19)
\cdot(10\cdot 18\cdot 20)
\\
&>2^{\displaystyle1+2+3+4+5+6+12+16+11} 
\\
&= 2^{\displaystyle60}\ .
\end{aligned}
$$
