$2^nn!\le n^n$ for $n\ge6$

Question

Guys I just can't solve this one. Really need some help with it.

Show that this inequality holds true:

1. $2^nn!\le n^n$ for all $n\ge6$
2. $n^n\le3^nn!$ for all $n\ge1$

Answer 1

Induction usually helps, for the first one, $n=6$ gives $$2^6\cdot 6! = 46080 \le 6^6=46656$$ so the base case is done, assume it is true for $n$, then for $n+1$ we have $$2^{n+1}(n+1)! = 2^n n! \cdot2(n+1)\leq n^n\cdot2(n+1)=2n^{n+1}+2n^n < (n+1)^{n+1}$$ where hte last inequality comes from binomial theorem, you can do similarly for the last one.

Answer 2

Hint $\left(\frac{n+1}{n}\right)^n$ is an increasing function of limit $e$ (and $2<e<3$)

If you can show that, then inductions are easy.