Prove the following inequalities,$$ (n!)^2 ≤ n^n(n)! <(2n)!$$
My attempt
I proved one of the inequality using mathematical induction.
To prove - $ (n!)^2 ≤ n^n(n)!$
For $ n = 1 $,
LHS $≤$ RHS.
Let us assume that for $ n = k$ the inequality is true.
$\therefore$ $(k!)^2 ≤ k^k(k)!$ –––––> [1]
Now let us prove that $ ((k+1)!)^2 ≤ (k+1)^{(k+1)}(k+1)!$ for $ n = k+1$.
Taking LHS,
$=>(k+1)^2(k!)^2 $
$≤(k+1)^2k^k(k)!$ –––––> [From [1]]
$≤(k+1)(k)^k(k+1)!$
From above we can deduce that,
$LHS≤RHS$.
How to prove the second inequality given in the question?
The left inequality is equivalent to $$ n! = 1\cdot 2 \ldots n \le n \cdot n \ldots n = n^n. $$
The right inequality is equivalent to $$ n^n = n \cdot n \ldots n < (n+1) \cdot (n+2) \ldots 2n = \frac{(2n)!}{n!}. $$