Prove, for all integers $n \geq 6$, $n^n>2^n n!$.

Question

I tried method of induction. Assuming its true for $n=k$, i.e. $k^k > 2^k k!$. To prove, $(k+1)^{(k+1)} > 2^{k+1} (k+1)!$. I started with, $$\begin{align}2^{k+1}(k+1)! &= 2^k 2 (k+1) k! \\ &< k^k 2 (k+1) \\ &< k^k k (k+1) \\ &= k^{(k+1)}(k+1) \\ &< (k+1)^{(k+1)} (k+1)\\ & = (k+1)^{(k+2)}\end{align}$$ but I need to end up at $(k+1)^{(k+1)}$.

Any thoughts?

Answer 1

For $n=6$ we need to prove that $$6^6>2^66!$$ or $$3^6>6!$$ or $$729>720.$$

If $n^n>2^nn!$ then we obtain: $$(n+1)^{n+1}=\left(1+\frac{1}{n}\right)^n(n+1)n^n>\left(1+\frac{1}{n}\right)^n(n+1)2^nn!.$$ Thus, it's enough to prove that $$\left(1+\frac{1}{n}\right)^n(n+1)2^nn!>2^{n+1}(n+1)!$$ or $$\left(1+\frac{1}{n}\right)^n>2,$$ which is obvious for $n\geq6$.

Answer 2

$$(n+1)^{n+1}>2^{n+1}(n+1)!$$

$$(n+1)^{n}>2^{n+1}(n)!$$

$$\frac 1 2 (n+1)^{n} >2^{n}(n)!$$

By hypothesis of induction we have that $n^n> 2^n n!$

$$\frac 1 2 (n+1)^{n} >n^n>2^{n}(n)!$$

$$( \frac {n+1}{n})^n>2$$

Which is true for $n>2$ since :

$$2\leq (1+\frac 1 n)^n<e$$