Which one is bigger $2^{n!}$ or $(2^{n})!$? Which one is bigger $2^{n!}$ or $(2^{n})!$ ? 
where $n\in\mathbb N$.
 A: It's enough to take this limit: $\lim_{n \rightarrow \infty} \frac{2^{n!}}{(2^n)!}$, by the Stirling formula we get $\lim_{n \rightarrow \infty} \frac{2^{(\frac{n}{e})^n \sqrt{2 \pi n}}}{(\frac{2^n}{e})^{2^n} \sqrt{2 \pi 2^n}}$, which is asymptotic (after elevating and taking logarithms) to $\lim_{n \rightarrow \infty} \frac{\exp(\frac{n^n}{e^n})}{\exp(2^n 2^{n/2})} = \lim_{n \rightarrow \infty} \frac{\exp(\frac{n^n}{e^n})}{\exp(2^{3/2 n})}$. Now, $\frac{n^n}{e^n}$ is bigger than $2^{3/2n}$ (and it is easy to verify). So the limit is infinite, which means that $2^{n!}$ is bigger than $(2^n)!$, if $n$ is big, of course.
Little note: I know that the asymptotic preseves constants and, even though I omitted them, it's good practice always to include them.
A: Actually, $(2^n)!$ is larger if 1 < n < 4.97399743597. See this graph for detail.
A: By proceeding straightforwardly we can answer without using Stirling's formula.
To start we show that $(n-1)!>2^n$ for large n. [by induction]
First note that $5!>2^6$.
Then we note that if we increase n by 1 we multiply the left hand side by n, and the right hand side by 2. Since n>2 we conclude that the left hand side remains larger for bigger and bigger n.
Multiplying by n we get $n!>n2^n$
Replacing n with $log(2^{n})$ [using base-2 logarithms] gets us: $n!>2^{n}log(2^{n})$
Note that $2^{n}log(2^{n})$=
$log(2^{n})$+$log(2^{n})$+$log(2^{n})$+...$log(2^{n})$+$log(2^{n})$
[$2^n$ terms]
$>log(2^{n})$+$log(2^{n}-1)$+$log(2^{n}-2)$+...log(2)+log(1) 
[since this has the same amount of terms, and some of them are smaller.]
=$log(2^{n}!)$ [combining logs]
Putting together these inequalities gives us that n!>$log(2^{n!})$, and raising 2 to these powers gives $2^{n!}$>$(2^{n}!)$ [for n>5 since our induction only told us that $(n-1)!>2^n$ for n>5]
A: We should expect $2^{n!}$ be larger. Note that $(2^n)! \leq (2^n)^{(2^n)} = 2^{(n 2^n)}$. 
So we want to prove $n 2^n < n!$ eventually. (And this can be done very simply.)
A: Indefinite:
for $n=0$: $2^{0!}=2^1=2,$ $(2^0)!=1!=1$, so $2^{0!}>(2^0)!$
for $n=2$: $2^{2!}=2^2=4,$ $(2^2)!=4!=12$, so $2^{2!}<(2^2)!$.
A: $ 2^{n!}=(2^n)^{(n-1)!} $, so we have to compare $(2^n)^{(n-1)!} $ and $ (2^n)! $. For $n\geq 6$, $(n-1)!>2^n$. I may be wrong, but doesn't the result follow immediately from this? It's still true for $n=5$ though, but not for $n=4$.
