Proof that $n^n<(n!)^2$ for $n>2$

Question

Prove that $n^n<(n!)^2$ for $n>2$

I tried math induction, but couldn't prove that $(k+1)^{k+1}<((k+1)!)^2$.

Answer 1

The inequality is equivalent to $$ \underbrace{n\cdot n \cdot\cdots\cdot n}_{n} \leq 1^2\cdot 2^2\cdot \cdots\cdot n^2=(1\cdot n)(2\cdot (n-1))\cdots (n\cdot 1) $$

Hence it remains to show that $n \le k(n+1-k)$ for $n\ge k \ge 1$.

$n \le k(n+1-k) \iff n \le kn+k-k^2 \iff (k-k^2)+(kn-n) \ge 0 \iff (k-n)(1-k) \ge 0$.

but $(k-n)(1-k) \ge 0$ iff $(k-n)(k-1) \le 0$, and this happens iff $k$ is between $1$ and $n$ inclusive. Hence we have proved our claim and solved the problem

Answer 2

Here's the inductive step:$$(n+1)^{n+1}<((n+1)!)^2 \\ (n+1)^{n+1}<n!\cdot n!\cdot(n+1)^2 \\ (n+1)^{n-1}<n!\cdot n!. $$ Hence it suffices to show $$(n+1)^{n-1}\le n^n,$$ and indeed it is equivalent to $$(n+1)^n\le n^n(n+1) \\ \left( \frac{n+1}{n}\right)^n=\left(1+\frac{1}{n}\right)^n\le n+1.$$ Since the LHS is bounded by $e$, the latter inequality holds for all $n\ge2,$ and by calculation we have equality for $n=1$. And thus we're done.

Edit suggested by @user161825:

In fact, for $n\ge 2$ we have that $\displaystyle \frac{(n+1)^{n-1}}{n^n}$ strictly decreases, hence we obtain $$\frac{n^n}{(n!)^2}\le \frac{3}{4} $$ for $n> 2$.

Answer 3

Hint: To do this you first prove: $(k+1)^k < k^{k+1}$ by induction.

Answer 4

Here is an argument based on Stirling's Approximation: $$ \frac{n^n}{(n!)^2}\leq \frac{e^n}{n!\sqrt{2\pi n}}\leq \frac{e^3}{6\sqrt{6\pi}}< 0.78. $$ This follows from the observation that the sequence $\left(\frac{e^n}{n!\sqrt{2\pi n}}\right)_{n\geq 3}$ is decreasing since $n\geq e$ in this range.