Which one is bigger $2^{n!}$ or $(2^{n})!$?

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.)

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.

Actually, $(2^n)!$ is larger if 1 < n < 4.97399743597. See this graph for detail.