Comparison between $a^{b!}$ and $(a^b)!$

Question

I've recently discovered two conclusions that may be true. However I can't prove it. The statement is as follows :

Suppose $a$, $b$ are positive integers and $a\geq b \geq 2$, then I found this conclusion $a^{b!}<(a^b)!$ might be true. On the other hand, for any $a\geq 2$, there must exist an integer $b_0>a$ ( $b_0$ is related to $a$ ) such that for any $b\geq b_0$ we have $a^{b!}>(a^b)!$ .

I have verified the above conclusion in some cases such as $2^{100!}>2^{100}!$ , $99^{280!}>99^{280}!$ , $100^{100!}<100^{100}!$ , $200^{100!}<200^{100}!$ .

For $a=2$ , we can verify that $2^{5!}>2^{5}!$ is true and $b_0=5$ is the smallest one.

Answer 1

Hint:

Stirling's approximation is what I would use if I care about these $a$ and $b$s being small since this inequality doesn't always hold, otherwise consider this:

1. Prove using induction that $n!<n^n$ when $n>2$
2. Conclude from it that: $$(a^b)!<a^{ba^b}<a^{b!}$$ for large enough $b$ (why does $b!$ grow faster than $ba^b$?).

Answer 2

Guessing your answer to Theo's question in the comments would be "yes", here is a proof for your "On the other hand, for any $a\geq 2$, there must exist an integer $b_0>a$ ( $b_0$ is related to $a$ ) such that for any $b\geq b_0$ we have $a^{b!}>(a^b)!$".

For $a\ge2$ fixed, as $b\to+\infty,$

$$\ln\ln(a^{b!})=\ln(b!)+\ln\ln a\sim\ln(b!)\sim b\ln b,$$ whereas $$\ln\ln((a^b)!)\sim\ln(a^b\ln(a^b))=b\ln a+\ln(b\ln a)\sim b\ln a$$ hence for $b$ large enough, $\ln\ln(a^{b!})<\ln\ln((a^b)!),$ i.e. $a^{b!}<(a^b)!.$