How would you go about finding the exponent for any given x that most closely matches or surpasses factorial growth?

Question

While analyzing the factorial function and comparing it to basic exponentiation, I couldn't help but notice the obvious fact that exponentiation can eventually overtake factorialization if the exponent is big enough. $200^{200}$ for example far surpasses $200!$.

But I'd be interested in finding the smallest exponent that equals or surpasses $200!$ - and likewise maybe also possibly also the largest one that's inferior. In other words, I'm looking for $a$ and $b$ such that $200^a < 200! \leq 200^b$.

And of course $200$ is just one possible value for $x$. So in more general terms, I'm looking for either a formula, or a list of steps that can be taken to find $a$ and $b$ such that:

$x^a < x! \leq x^b$, for any $x$.

Answer 1

In virtue of Stirling's formula, stating that when $n$ goes to infinity, $$ n! \sim \sqrt{2\pi n}e^{-n}n^n $$ one has that for any $a >0$, $$ \frac{n!}{n^a} \sim \sqrt{2\pi n}e^{-n} n^{n-a}, $$ which goes to $+\infty$ with $n$, and is in particular not bounded above. Therefore, there is no $a$ such that $n! \leqslant n^a$ for all $n$.

Answer 2

Use algebra to find the solution of $$n!=n^a \quad \implies \quad\log(n!)=a \log(n)\quad \implies \quad \large\color{red}{a=\frac {\log(n!)}{\log(n)}}$$

Assuming that $n$ is large, use Stirling approximation $$\log(n!)=n (\log (n)-1)+\log \left(\sqrt{2 \pi n } \right)+\frac{1}{12 n}-\frac{1}{360 n^3}+O\left(\frac{1}{n^5}\right)$$

For $n=200$, the above gives $a=162.926$.

Checking $$200!- 200^{162}=7.828\times 10^{374}$$ $$200^{163}-200!=3.805\times 10^{374}$$

$$200^{162} < 200! <200^{163}$$