Which is larger? or

I often suggest this as a problem to be solved by completely elementary means: no Stirling, no logs, no nothing.

The key observation is that n! < n^n. This should be obvious: n! is the product of n numbers ranging from 1 to n, while n^n is the product of n numbers which are all n. As soon as n is at least 2, the factorial loses badly to the exponent of a number by itself.

Therefore,

\displaystyle \left(2^{100}\right)! < \left(2^{100}\right)^{2^{100}}

We now have a little tower of exponents, which we can simplify:

\displaystyle \left(2^{100}\right)^{2^{100}} = 2^{100 \cdot 2^{100}}

We wish to further simplify the exponent here. Clearly, 100 is less than 128 which is 2^7, so

\displaystyle \left(2^{100}\right)! < 2^{2^{107}}

This is great, because the other number we need to compare this with is also a power of 2: it is 2^{100!}. The only thing left to do is confirm that

\displaystyle 2^{107} < 100!

It’s useful to know how many times factorials are divisible by primes: for example, when calculating 100!, we’re multiplying together 50 even numbers, of which 25 are multiples of 4, of which 12 are multiples of 8, of which 6 are multiples of 16, of which 3 are multiples of 32, and there’s even one multiple of 64 (which is 64 itself, of course). This gives us

2^{50+25+12+6+3+1}=2^{97}

So 100! is actually divisible by 2^{97}, and besides that it’s also obviously divisible by 11,13,17 and 19 whose product is more than 10000, so it must be way larger than 2^{97}\cdot 10000 which is way more than 2^{97}\cdot 2^{10}.

So, all told, 2^{100!} is much, much larger than \left(2^{100}\right)!.