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.

I know I am beating a dead horse here, but anyways. If you can accept some "handwaving" this is an argument a normal higschool student should be able to accept.
We have that if $a>b$ then $\log a > \log b$. Now when comparing big numbers, it is often easier to look at the logarithms. The crux here is to note that 
$$
  \log 20! > 20 \log(20)-20
$$
Let us assume that $20!<2^{40}$, and see if we obtain a contradiction.
\begin{align*}
    \log 2^{40} & > 20 \log(20)-20\\ 
    40 \log 2 & > 20(\log 5 + 2\log2)-20\\
     0 & > 20 \log 5 - 20    \\ 
       1 & > \log 5
\end{align*}
Which is rubbish since $\log 5 > \log e = 1$. Therefore
$$
   20! > 2^{40}
$$
To see that
$$
   \log n! > n \log n - n \ , \quad \forall \, n>0
$$
One can note that 
$$
    \log n! = \sum_{k=1}^n \log k > \int_1^n \log x\,\mathrm{d}x = n \log n - n
$$
Where the inequality comes from the following figure

