Which is larger? $20!$ or $2^{40}$?

It is probably easier to note that $2^{40} = 4^{20}$. The only ones of the 20 factors in $20!$ that are smaller than $4$ are $1$, $2$ and $3$. But, on the other hand, $18$, $19$ and $20$ are all larger than $4^2$, so we can see $$ 20! = 1\cdot 2\cdot3\cdots 18\cdot19\cdot 20 > \underbrace{1\cdot1\cdot1}_{3\text{ ones}}\cdot\underbrace{4\cdot4\cdots 4\cdot 4}_{14\text{ fours}}\cdot\underbrace{16\cdot16\cdot 16}_{3\text{ sixteens}} = 2^{40} $$ by comparing factor by factor.

You can simply compute both numbers and compare them: $$ 20! = 2432902008176640000 \\ 2^{40} = 1099511627776 $$

So you can easily see that $2^{40} < 20!$

Use a multiplicative variant of Gauss's trick: $$ (n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1) \ge n^n $$ Setting $n=20$, we get $20!\ge 20^{10} > 16^{10} = 2^{40}$.