Inequality and induction. I can't solve this inequality by use of induction,
$$(n!)^2\ge n^n$$
so my teacher has advised me to break the multiplies into pairs, but it doesn't help. How do I prove it?
 A: $(n!)^2 = (1*2*.....*n)(1*2*.....*n)$
$ = (1*2*.....*n)(n*(n-1)*.....*1)$
$ = (n*1)((n-1)*2)*.....*((n+1-k)*k)*....(1*n)$
If we can prove $(n+1-k)*k \ge n$ then we are done as 
$(n*1)((n-1)*2)*.....*((n+1-k)*k)*....(1*n) \ge (n)(n)(n).....(n) = n^n$
.....
And $(n+1-k)*k = nk + (1-k)k = n + (k-1)n -(k-1)k = n + (k-1)(n-k)$.  As $0 \le k \le n$ then $(k-1)(n-k) \ge 0$ and $(n+1-k)*k \ge n$.
A: I'll try to continue @Dr. Sonnhard Graubner proof by showing that 
$$n+1\geq (1+\frac{1}{n})^{n} \forall n\geq1$$
Let $f(x)=ln(x+1)-xln(1+\frac{1}{x})$ then $f'(x)=\dfrac{2}{n+1}-ln(1+\frac{1}{n})$ and $f''(x)=\dfrac{1-x}{x(x+1)^2}<0$ for $x>1 $ so $f'$ is decreasing and $$\lim_{x\to \infty}f'(x)=0$$
so $f'(x)>0$ for $x>1\Rightarrow f(x)\geq f(1)=0$ for $x\geq1$ which is what we wanted
A: This is  a supplement to @fleablood's nice answer. Note that the advice of OP's teacher to  break the multiples into pairs allows a proof even without induction.

We obtain for $n\geq 1$
  \begin{align*}
\color{blue}{\left(n!\right)^2}=\prod_{k=1}^n (k\cdot k)=\prod_{k=1}^n \left[k\cdot (n+1-k)\right]
=\prod_{k=1}^n[n+\underbrace{(k-1)(n-k)}_{\geq 0}]\color{blue}{\geq n^n}
\end{align*}
  and the claim follows.

A: you have to prove that $$(n+1)!^2\geq (n+1)^{n+1}$$ if $$(n!)^2\geq n^n$$
we write the last inequality in the form $$n!\geq \sqrt{n^n}$$ we multiply by $$n+1>0$$ and get
$$(n+1)!\geq (n+1)\sqrt{n^n}$$ and now we Show that
$$(n+1)\sqrt{n^n}\geq \sqrt{(n+1)^{n+1}}$$ and this is
$$(n+1)^{2}n^{n}\geq (n+1)^{n+1}$$ or
$$(n+1)\geq \left(\frac{n+1}{n}\right)^{n}$$ which is 
$$n^n\geq (n+1)^{n-1}$$
and his is $$n+1\geq \left(1+\frac{1}{n}\right)^n$$
A: For a generic $n$, if it is true that
$$
n^{\,n}  \le \left( {n!} \right)^{\,2}  = \left( {\prod\limits_{1\, \le \,k\, \le \,n} k } \right)^{\,2}  = \prod\limits_{1\, \le \,k\, \le \,n} {k^{\,2} } 
$$
it means that it is true that
$$
1 \le \prod\limits_{1\, \le \,k\, \le \,n} {k{k \over n}}  = P(n)
$$
then, since
$$
\eqalign{
  & {{P(n + 1)} \over {P(n)}} = {{\prod\limits_{1\, \le \,k\, \le \,n + 1} {k{k \over {n + 1}}} } \over {\prod\limits_{1\, \le \,k\, \le \,n} {k{k \over n}} }} 
= {{\left( {n + 1} \right)\prod\limits_{1\, \le \,k\, \le \,n} {k{k \over {n + 1}}} } \over {\prod\limits_{1\, \le \,k\, \le \,n} {k{k \over n}} }} =   \cr 
  &  = \left( {n + 1} \right)\prod\limits_{1\, \le \,k\, \le \,n} {{n \over {n + 1}}}  = \left( {n + 1} \right)\left( {{1 \over {1 + 1/n}}} \right)^{\,n}  
= {{\left( {n + 1} \right)} \over {\left( {1 + 1/n} \right)^{\,n} }} \ge {{\left( {n + 1} \right)} \over e} \ge {{\left( {n + 1} \right)} \over 3} \cr} 
$$
we have that
$$
P(n) \le P(n + 1)\quad \left| {\;2 \le n} \right.
$$
Now, it is
$$
\left\{ \matrix{
  1 \le P(0) = {{\left( {0!} \right)^{\,2} } \over {0^{\,0} }} = 1 \hfill \cr 
  1 \le P(1) = {{\left( {1!} \right)^{\,2} } \over {1^{\,1} }} = 1 \hfill \cr 
  1 \le P(2) = {{\left( {2!} \right)^{\,2} } \over {2^{\,2} }} = 1 \hfill \cr}  \right.
$$
and therefore we can conclude that
$$
0 \le P(n) = {{\left( {n!} \right)^{\,2} } \over {n^{\,n} }}\quad \left| {\;0 \le n} \right.
$$
