How to prove $\int_0^\infty e^{-x}\ln^n \frac{1}{x}dx<n!$?

Question

Once I met a problem about limits: $$\lim_{n\rightarrow\infty}\int_0^\infty e^{-x}\frac{\ln^n \frac{1}{x}}{n!} dx=1$$
After I proved it, I used Mathematica and discovered that the sequence $\int_0^\infty e^{-x}\frac{\ln^n\frac{1}{x}}{n!} dx$ seems to be monotone increasing to its supremum $1$.
I guess so, but I’m confused about how to prove.
My question is: How to prove that $$\int_0^\infty e^{-x}\ln^n \frac{1}{x}dx<n!,\quad \forall n\ge 1$$
and the monotonity of $$\int_0^\infty e^{-x}\frac{\ln^n\frac{1}{x}}{n!} dx$$

Additional: I have already proved that $$\lim_{n\rightarrow\infty} \frac{2^{n+1}}{n!}\left(n!-\int_0^\infty e^{-x}\ln^n \frac{1}{x}dx\right)=\lim_{n\rightarrow\infty} \frac{2^{n+1}}{n!}\left(\int_0^1 \left(1-e^{-x}\right)\ln^n\frac{1}{x} dx-\int_1^\infty e^{-x}\ln^n\frac{1}{x} dx\right)=1$$
by using the inequality $x-\frac{x^2}{2}\le 1-e^{-x}\le x$ and $\ln^n x<n! x\left(x\ge 1\right)$.
It can indicate that $1-\int_0^\infty e^{-x}\frac{\ln^n \frac{1}{x}}{n!} dx$ is EVENTUALLY (which means for sufficient large $n$) positive and monotone decreasing at $O\left(2^{-n}\right)$.

Answer 1

I think you can go with this walk:

Let be $f(x) = e^{-x} \ln^n{\frac{1}{x}}$ consider the function $g(x) = e^{-x}x^n$.

We know that $ \int_{0}^{\infty} g(x) dx = \int_{0}^{\infty}e^{-x}x^n dx = \Gamma(n+1) = n!$

$f(x) < g(x) \Rightarrow \int_{0}^{\infty} f(x) dx < \int_{0}^{\infty} g(x) dx = n!$

for proving this, we need to prove that $\ln^n \frac{1}{x}< x^n$. So, let be $h(x) = x^n - \ln ^n {\frac{1}{x}}$

$h'(x)> 0 \Rightarrow n x^{n-1} - (n \ln^{n-1}(\frac{1}{x}))\frac{1}{\frac{1}{x}}(-\frac{1}{x^2}) = n(x^{n-1} + \frac{1}{x}\ln^{n-1}(\frac{1}{x})) > 0 \Rightarrow x^n > (-1)^{n-1} \ln ^{n-1}x$

I think it done.