I'm having trouble in solving this problem:
Prove by induction $(2 n)! \ge (n!)^2, \forall\ n\ge 0$
So far I came up with this:
$P(0): 2\ge1$ true
$P(n)\implies P(n+1): (2n+2)!\ge?[(n+1)!]^2$
$(2n+2)!=(2n)!\cdot(2n+2)\ge(n!)^2\cdot(2n+2)=$
$=(n!\cdot \frac{n+1}{n+1})^2\cdot(2n+2)=[(n+1)!\cdot \frac1{n+1}]^2\cdot2\cdot(n+1)=$
$=[(n+1)!]^2\cdot\frac{2\cdot(n+1)}{(n+1)^2}=[(n+1)!]^2\cdot\frac2{n+1}$
$$[(n+1)!]^2\cdot\frac2{n+1}\ge?[(n+1)!]^2$$
Now, this shouldn't be true anymore because if $n>1$ then $\frac2{n+1}<1$. Do you have any suggestion? I can't handle this kind of problems very well yet.
Firstly, it should be $$P(1): 2\geq1.$$ Also, $P(0):1\geq1$ and by assumption of the induction we obtain: $$(2n+2)!=(2n)!(2n+2)(2n+1)>(n!)^2(n+1)^2=((n+1)!)^2.$$