which is bigger $2^{n^{1.001}}$ or $n!$ for a big enough n, How to detrmine which is bigger?
$2^{n^{1.001}}$ or $n!$
I have tried to make a series:  $a_n = \frac{2^{n^{1.001}}}{n!}$
and then try finding the limit of $\frac{a_{n+1}}{a_n}$
and see if its bigger or smaller then one, also tried to use induction but didnt really got my anywhere
btw I know the answer is that $2^{n^{1.001}}$ is bigger
 A: Why didn't it lead you anywhere?
\begin{aligned}
\frac{a_{n+1}}{a_n} &= \frac{\frac{2^{(n+1)^{1.001}}}{(n+1)!}}{\frac{2^{n^{1.001}}}{n!}} \\
                    &= \frac{2^{(n+1)^{1.001}}}{(n+1)2^{n^{1.001}}}
\end{aligned}
Take the log of both sides.
\begin{aligned}
\log\left(\frac{a_{n+1}}{a_n}\right) &= \log\left(\frac{2^{(n+1)^{1.001}}}{(n+1)2^{n^{1.001}}}\right) \\
&= \log\left(2^{(n+1)^{1.001}}\right) -\log\left(2^{n^{1.001}}\right) - \log\left(n+1\right) \\
&= (n+1)^{1.001}\log\left(2\right) - n^{1.001}\log\left(2\right) - \log(n + 1)
\end{aligned}
Divide both sides by $\log(2)$.
\begin{aligned}
\log_2\left(\frac{a_{n+1}}{a_n}\right) &= (n+1)^{1.001} - n^{1.001} - \log_2(n + 1)
\end{aligned}
Now, we have a new function $f(n)$ for which the following is true:
$$
f(n) = (n+1)^{1.001} - n^{1.001} - \log_2(n + 1)
$$
$$
\lim_{n \to \infty}f(n) > 0 \implies 2^{n^{1.001}} > n!\\
\lim_{n \to \infty}f(n) < 0 \implies 2^{n^{1.001}} < n!
$$
Let's break up this function into two parts:
$$
f(n) = (n+1)^{1.001} - \left(n^{1.001} + \log_2(n + 1)\right)
$$
Now, let's repeat the process:
$$
\lim_{n \to \infty} \frac{(n+1)^{1.001}}{\left(n^{1.001} + \log_2(n + 1)\right)} > 1 \implies \lim_{n \to \infty}f(n) > 0
$$
Here, we can simply say that the highest degree term of the numerator is larger than that of the denominator, so the limit goes to $\infty$. Therefore, following the chain of implications back up, we end up with $2^{n^{1.001}} > n!$. If you followed that train of thought, you could indeed come to the answer.
A: Hint: $$n!\le n^n=2^{n\log_2 n}$$
Spoiler:

 $2^{n\log_2 n}<2^{n^{1.001}}\iff n\log_2 n<n^{1.001}\iff \log_2n<n^{0.001}$, which is true eventually as $\log_2 n$ grows slower than any $n^\alpha$ with $\alpha>0$. (e.g. by L'Hôpital's rule on $n^{\alpha}/\log_2 n$)

A: Hint: Compare
$$
n^{1.001}\log(2)
$$
and the log of Stirling's Formula
$$
n\log(n)-n+\frac12\log(2\pi n)
$$

Note that since $x\gt\log(x)$ for all $x\gt0$, we have
$$
\begin{align}
n^{0.001}
&=\color{#C00000}{n^{0.0005}}n^{0.0005}\\
&\ge\color{#C00000}{0.0005\log(n)}n^{0.0005}
\end{align}
$$
Therefore, for $n\gt1$,
$$
\begin{align}
\frac{n^{0.001}}{\log(n)}
&\stackrel{\hphantom{n\to\infty}}{\ge}0.0005n^{0.0005}\\
&\stackrel{n\to\infty}{\to}\infty
\end{align}
$$
