Growth rate of $l(n)^n$ Let $l(n) = \alpha n + \beta$ be an affine function, where $\alpha,\beta\in\mathbb{Q}$ and $\alpha>0$. I have two questions with respect to the rate of growth of $l(n)^n$:

*

*Is it true that $l(n)^n\geq n!$ for $n\gg 0$?

*Is it possible to find bounds (even horrible ones) that only depend on $\alpha$ and $\beta$ for the previous inequality to be true?

Besides the obvious cases (e.g. $\alpha=1$) I'm not able to find an explicit bound and the computer is bad at handling these numbers. Any help is warmly welcome!
 A: $$(\alpha n+\beta)^n > n! \implies \Delta_n=n \log(\alpha n+\beta) - \log(n!) >0$$
As other users commented and answered, using Taylor series and Stirling approximation for large values of $n$, we end with
$$\Delta_n= (1+\log (\alpha ))\,n+\left(\frac{\beta }{\alpha }-\frac{1}{2} \log (2 \pi 
   n)\right)-\frac{\alpha ^2+6 \beta ^2}{12 \alpha ^2 n}+\frac{\beta ^3}{3 \alpha ^3 n^2}+O\left(\frac{1}{n^3}\right)$$ and then @K.defaoite's conclusions.
If $\alpha \geq 1$, this gives as bounds
$$\frac{e^{\frac{\beta }{\alpha }}}{\sqrt{2 \pi }} n^{\log (\alpha \sqrt e)}\exp\left(-\frac{\alpha ^2+6 \beta ^2}{12 \alpha ^2 n} \right)<\frac{(\alpha n+\beta)^n } {n!} <\frac{e^{\frac{\beta }{\alpha }}}{\sqrt{2 \pi }} n^{\log (\alpha\sqrt e )}$$
One the other side,we can approximate the value of $n$ which would make $\Delta=0$. This is given by
$$n_*=-\frac{1}{2 \log (\alpha e)}W\left(-\frac{\log (\alpha e )}{\pi
   }e^{\frac{2 \beta }{\alpha }}\right)$$ which would be a real if the argument of Lambert function $W(.)$ is $\leq - \frac 1e$
A: I'm going to use $a$ and $b$ instead of $\alpha$ and $\beta$ for convenience. Let $\ell(n|(a,b))=an + b$. We want to investigate whether $\ell(n)^n \succeq n!$ in general. I'm going to use Stirling's approximation, namely
$$\sqrt{2\pi n}\left(\frac{1}{e}n\right)^n \asymp n!$$
First observation: $\ell(n)^n \asymp (an)^n$. We can rewrite
$\ell(n)=n(a+b/n)$ which tends to $an$ as $n$ gets large. So now we want to investigate if
$$(an)^n \succeq \sqrt{2\pi n}\left(\frac{1}{e}n\right)^n$$
Let's begin. Call the left side $p(n)$ and the right side $q(n)$. Let's first assume $n\in \mathbb{R}_{\geq 0}$ and assume that both sides are continuous (this is true, but requires some care to rigorously prove). First, a simple observation. Given two continuous functions $f$ and $g$ on $\mathbb{R}$, if $\exists x_* \in \mathbb{R} \text{ such that } f'(x)>g'(x) \forall x>x_*$, then $\exists x_{**} \text{ such that } f(x)>g(x) \forall x>x_{**}$. Now I'll use the following identity:
$$\frac{\mathrm{d}}{\mathrm{d} x}(f(x)^{g(x)})=f(x)^{g(x)}(g'(x)\ln(f(x))+\frac{f'(x)}{f(x)}g(x))$$
Some algebra shows us that
$$p'(n) = (an)^n(\ln(an)+1)$$
And
$$q'(n)=\left(\frac{1}{e}n\right)^n\left(\sqrt{\frac{\pi}{2n}}+ \sqrt{2\pi n}\left(\ln\left(\frac{1}{e}n\right)+1\right)\right)$$
Since $q'(n) > p'(n)$ if $a\leq\frac{1}{e}$, we can say that $\ell(n)^n \succeq n!$ is definitely false if $a\leq\frac{1}{e}$, definitely true if $a\geq 1$(this isn't all that difficult to prove, but I can illustrate it if you like) but for values of $a$ in the range $\left(\frac{1}{e},1\right)$ more sophisticated analysis is needed. I'll give it a try later, but I can't make any guarantees I'll come up with anything conclusive.
