$1 < \log n < n < n\log n < n^2 < n^3< \ldots < 2^n < 3^n < 4^n < \ldots < n! < n^n$
I am trying to find out if $n!$ is greater than $4^{n}$. One way to attempt to solve this is to compare $n!$ with $4^{n}$ by taking a limit as $n$ approaches infinity, and for that, I would use the limit:
$$ \lim_{z\to \infty} \frac{n!}{4^{n}} $$
But to solve this limit, I needed to use L'Hôpital's Rule, by taking the derivative of both the numerator and the denominator.
However, I cannot directly differentiate $n!$ since it is a discrete function. To handle this, I used the Gamma function, which is a continuous function such that $\Gamma(n-1) = n!$
$$ \lim_{n\to \infty}\frac{\int_{0}^{\infty }t^{n-1}\cdot e^{-t}dt}{4^{n}} $$
However, I am having trouble finding an answer that indicates whether $n!$ grows faster than $4^{n}$ or not.
Could you help me? I'm not sure what strategy to use
Use induction on $n$. Prove the fact for some $n > 4$. Then $$4^{n+1} = 4\cdot 4^n < 4n! < (n+1)n! = (n+1)!$$