Limit of factorial how to continue

Question

$$\lim_{n\to\infty}\left(\dfrac{(n+1)^{n+1}\cdot n!}{2^{(n+1)!-n!}}\right)=\lim_{n\to\infty}\left(\dfrac{(n+1)^{n+1}\cdot n!}{2^{n!\cdot n}}\right).$$

How to continue?
the answer is $0$ ... thank you

Answer 1

If you take the natural log we get

$$ \ln(a_n)=(n+1)\ln(n+1)+\ln(n!)-( (n+1)!-n! )\ln(2)$$

$$= (n+1)\ln(n+1)+\sum_{1}^{n}\ln(k)-n\,n!\ln(2) $$

$$ = (n+1)\ln(n+1)+n\ln(n)-n+1-n\,n!\ln(2) \sim_{ \infty} 2n\ln(n)-n-nn!\,\ln(2) $$

$$ = 2n\ln(n)-(n+1)!\,\ln(2) \to -\infty $$

$$ \implies a_n \rightarrow_{n\to\infty} 0 $$

Note:

$$ \sum_{1}^{n}\ln(k) \sim \int_{1}^{n} \ln(x) dx =n\ln(n)+n-1. $$