<s>[INST] 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 $2^{42}$.
 [/INST] $$20!=2^{18}\cdot3^8\cdot 5^4\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19$$
$$19\cdot 17\cdot 13=4199>2^{12}, \\3\cdot 11>2^5$$
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: $$\frac{20!}{2^{60}}=\left(1+\frac{103}{4096}\right)
\left(1+\frac1{32}\right)\left(1+\frac3{32}\right)^2\left(1+\frac18\right)^3\left(1+\frac{11}{64}\right)$$
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