algebra precalculus - Show that for any real $s geq 0$, $e^{-sqrt{n!}}(n!)^s < frac{1}{n^2}$ holds for sufficiently large $n$

Let $s \geq 0$ be a real number. Show that there exists a natural number $n_0 \in \mathbb{N}$ such that $\forall n \in \mathbb{N},n \geq n_0,$ we have
$$e^{-\sqrt{n!}}(n!)^s < \frac{1}{n^2}$$

By taking logarithms, we need to show that
$$-\sqrt{n!} + s \log{n!} + 2\log{n} < 0$$

for sufficiently large $n$.
It would be sufficient to show that the limit of the sequence is a negative number; In order to do that, I thought of using the Stirling approximation, $n! \sim \sqrt{2\pi n}\left( \frac{n}{e} \right)^n:$ then it would suffice to show
$$\lim_{n \to \infty} -\sqrt[4]{2\pi n}\left( \frac{n}{e} \right)^\frac{n}{2} + s\left(\frac{1}{2}(\log(2\pi) + \frac{1}{2}\log(n) + n(\log(n) - 1)\right)+ 2\log{n} < 0,$$
however this looks kind of scary.

Intuitively, $\sqrt{n!}$ is a polynomial in $n!$ and so clearly grows much faster than any multiple of $\log{n!}.$ It also clearly exceeds $2\log{n}$, but I don't know how to make this rigorous.

Any ideas?

Answer

It is easy to check that for fixed $s\geq 0$,

$$\lim_{x\to\infty}e^{-\sqrt{x}}x^{2s}=0$$
Therefore, if $x$ is sufficiently large,
$$e^{-\sqrt{x}}x^{2s}\leq 1\Longrightarrow e^{-\sqrt{x}}x^s\leq\frac{1}{x^s}$$
In particular, if $n$ is sufficiently large
$$e^{-\sqrt{n!}}(n!)^s\leq\frac{1}{(n!)^s}$$
and it is not hard to check that for sufficiently large $n$ we also have
$$\frac{1}{(n!)^s}\leq\frac{1}{n^2}$$