- #1

mikeyBoy83

## Homework Statement

I need to prove by induction that ##(n!)^{2} \le (2n)!##. I'm pretty sure about my preliminary work, but I just need some suggestions for the end.

## Homework Equations

It is well known from a theorem that if ##a \le b## and ##c \ge 0##, then ##ca \le cb##.

## The Attempt at a Solution

ATTEMPT AT PROOF: For ##n=1##, we can see that ##(1!)^{2} \le (2\cdot 1)!## holds. Now, assume that ##(k!)^{2} \le (2k)!## holds for ##k \in N##. Then it must be shown that ##n=k+1## holds. For that,

##[(k+1)!]^{2} = (k+1)^{2}k!^{2}## and by assumption ##(k!)^{2} \le (2k)!## implies with the theorem above that

##(k+1)^{2}k!^{2} \le (k+1)^{2}(2k)!##

And this is the point at which I don't know how to proceed. I need to conclude that this is in fact less than ##[2(k+1)]!##, but I am unsure how to do that. I would appreciate any direction.