proving that $n^n \le (n!)^2$

Question

I want to prove that $n^n \le (n!)^2$.

Now I tried by induction:

for $n=1$ ,$1=1$ and $P(1)$ is true

I suppose that $P(n)$ is true and I have to demonstrate that $P(n+1)$ is true $$((n+1)!)^2=(n+1)^2*(n!)^2 \le (n+1)^2*n^n=(n+1)(n+1)*n^n=(n+1)^n$$ But I'm not sure about the last passage

In addition to the last equation being obviously false, the inequality you already have points the wrong way, you want to prove that $((n+1)!)^2$ is smaller than something, but what you have shows it is larger than something. — Henrik supports the community, Mar 09 '18 at 16:48
Your proof is still attempting to go there wrong way. That needs to be a $\ge$ not a $\le$ symbol. And so you must prove $(n+1)(n+1)*n^n \ge (n+1)^{n+1}$ ... which might be tough. — fleablood, Mar 09 '18 at 17:23

fleablood · Answer 1 · 2018-03-09T17:41:54.680

Notice: $(n!) = (1*2*...... *n)^2 = (1*2*...... *n)*(1*2*...... *n)=$

$(1*n) * (2*(n-1)) * ....... * (n*1)$.

If we can prove that $k*(n-k+1) \ge n$ we will have

$(1*n) * (2*(n-1)) * ....... * (n*1) \ge n*n*n*....* = n^n$, and that will be it.

We can probably prove $k*(n-k+1) \ge n$ by induction but I'd rather note:

$k \le n$ and $k\ge 1$ so $k(n-k + 1) = kn - k^2 + k = n + (k-1)n -k^2 + k \ge n + (k-1)k - k^2 +k = n$.

.....

Intuitively this is also a result of AM-GM. $k$ and $n-k + 1$ average to $\frac {n+1}2$ so if $m = \frac {n+1}2$ and $k = m - d$ and $n-k+1 = m + d$ we have $k(n-k + 1) = (m - d)(m+d) = m^2 - d^2 \ge m^2 = \frac {(n+1)^2}4$. If $n\ge 3$ then $\frac {(n+1)^2}4 \ge \frac {(n+1)^2}{n+1} = n+ 1> n$. And for $n=1,2,3$ then $\frac {(n+1)^2}4 = 1, \frac 94, 4 \ge n$.

score 2 · Answer 2 · answered Mar 09 '18 at 16:47

What you are trying to prove is not true. For large enough $n$ (which means $\ge 3$) it will go the other way. If we just use Stirling's approximation $$(n!)^2 \approx \left(\frac {n^n}{e^n}\sqrt{2\pi n}\right)^2=\frac {n^{2n}}{e^{2n}}2\pi n$$ which will clearly be greater than $n^n$ when $n \gt e^2$. We can use one factor of $n^n$ to be greater than $e^{2n}$ and one to be equal to $n^n$

user · Accepted Answer · 2018-03-09T17:13:35.170

Before editing for $n^n \ge (n!)^2$

Note that it is false indeed

$$n^n=\overbrace{n\cdot n \cdot n\cdot...\cdot n} ^{n \, terms} $$

$$(n!)^2=(\overbrace{n\cdot (n-1) \cdot (n-2)\cdot...\cdot 1}^{n \, terms})^2\ge(n\cdot (n-1) \cdot (n-2)\cdot...\cdot 1)\cdot n\ge n^n$$

After editing for $n^n \le (n!)^2$

induction step

$$(n+1)^{n+1}=n^n(n+1)\frac{(n+1)^n}{n^n}=n^n(n+1)\left(1+\frac1n\right)^n\stackrel{Ind. Hyp.}\le(n!)^2(n+1)e \le ((n+1)!)^2$$

which is true for $n\ge 2$.

score 0 · Answer 4 · answered Mar 09 '18 at 17:07

0

This is false for $n \ge 3$. $27=3^3 < (3!)^2=36$.

Suppose $n^n <(n!)^2$ and $n \ge 3$. Then $\dfrac{(n+1)^{n+1}/((n+1)!^2)}{n^n/(n!)^2} =(n+1)(1+1/n)^n/(n+1)^2 \lt e/(n+1) \lt 1 $.

This shows that $n^n/(n!)^2$ is a decreasing sequence.

answered Mar 09 '18 at 17:07

marty cohen

107,799

proving that $n^n \le (n!)^2$

4 Answers4