Which is greater: $a^n! $ or $a^{n!}$? Out of pure curiosity, I have thought about the following:
"Let $a > 1$ and $n \in \mathbb{N}$. Which is greater: $a^n! $ or $a^{n!}$ ? And how do you prove it?"
Does anyone know a good/simple proof?
 A: Depends on the relationship between $a$ and $n$.
For all $a>1$, $a^{n!}$ will eventually outgrow $a^n!$ as $n \to \infty$. In general, $k! \le k^k$, so $(a^n)! \le (a^n)^{a^n} = a^{n \cdot a^n}$, and the exponent $n \cdot a^n$ grows slower than the exponent $n!$.
On the other hand, if $a$ is large and $n$ is small, then things can go the other way. You really want Stirling's approximation for factorials to say for sure.

Some more detail:
First, taking $n=a$ (when $a$ is reasonably large) is not large enough. Then $(a^a)!$ is a product of $a^a$ terms, most of which are close to $a^a$ in order of magnitude. On the other hand, $a^{a!}$ is a product of $a!$ (fewer) terms, which are only around $a$ (and therefore smaller).
For more accuracy, if we approximate $n! \approx (\frac ne)^n$ and compare it to $n a^n$ as we do in the argument above, then we see that we want $a < \frac ne$ or $n > e \cdot a$. Just taking $n = \lceil e a\rceil$ doesn't always work, since we took some approximations, but that tells us about how big $n$ has to be.
A: Since $k!<k^k$ if $k\ge2$,$$(a^n)!<a^{na^n}<a^{n!}$$for sufficiently large $n$.
