Evaluate the limit $\lim_{n\to \infty}{\frac{(n+3)!}{n^n}}$

Question

Evaluate the limit $$\lim_{n\to \infty}{\frac{(n+3)!}{n^n}}, n\in \mathbb N$$ I know that the limit is $0$ but how to prove it?

Answer 1

Note that $$\frac{(n+3)!}{n^n}=2\cdot3\cdot\frac{4\cdot\ldots\cdot n\cdot(n+1)\cdot(n+2)\cdot(n+3)}{n\cdot\ldots\cdot n}\le 6\cdot\frac4n\cdot\frac{n^{n-4}}{n^{n-4}}\cdot\frac{(2n)^3}{n^3}=6\cdot\frac4n\cdot8$$