Which is larger? $20!$ or $2^{40}$? Someone asked me this question, and it bothers the hell out of me that I can't prove either way.

I've sort of come to the conclusion that 20! must be larger, because it has 36 prime factors, some of which are significantly larger than 2, whereas $2^{40}$ has only factors of 2.

Is there a way for me to formulate a proper, definitive answer from this?

Thanks in advance for any tips. I'm not really a huge proof-monster.
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
$$ 20! = 1\cdot 2\cdot3\cdots 18\cdot19\cdot 20 &gt; \underbrace{1\cdot1\cdot1}_{3\text{ ones}}\cdot\underbrace{4\cdot4\cdots 4\cdot 4}_{14\text{ fours}}\cdot\underbrace{16\cdot16\cdot 16}_{3\text{ sixteens}} = 2^{40} $$
by comparing factor by factor.
With Stirling's formula:

$$ 4^n \lt \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \le n! \le e \sqrt{n} \left(\frac{n}{e}\right)^n $$

Both $\le$ are valid for $n\ge1$. Because $e\le 3$ the left inequality is valid a least for all $n \gt 12$.