Proving by induction that $(n^2)!>(n!)^2$ for $n \geq 2$

Question

I'm trying to prove that $(n^2)!>(n!)^2$ for $n \in [2,\infty) \cap\mathbb{Z^+}.$

Ok, here's what I've tried: $n \geq 2,$ $(n^2)!>(n!)^2$

$\color{green}{(2^2)!}=4!=24\color{green}{>(2!)^2}=4$, so the result is true in the base case.

Now assume that $(k^2)!>(k!)^2$.

Then, for $n=k+1$:

$[(k+1)^2]!=[k^2+2k+1]!=(k^2+2k+1)(k^2+2k)(k^2+2k-1)\cdots (k^2+1)[(k^2)!]>(k^2+2k+1)(k^2+2k)(k^2+2k-1)\cdots (k^2+1)\underbrace{(k!)^2}_{\text{induction hypothesis}}>\text{something I'm missing}>[(k+1)!]^2$

Could someone bridge (or at least hint towards) the gap?

Incidentally, how would one go about proving this result for all $n\in[2,\infty)$?

Thanks

Answer 1

All you need is $$\eqalign{[(k+1)^2]! &=(k^2+2k+1)(k^2+2k)\cdots(k^2+1)(k^2)!\cr &>(k+1)^2(1)\cdots(1)(k!)^2\cr &=[(k+1)!]^2\ .\cr}$$

On the other hand, do you have to do it by induction? If not then $$\eqalign{(n^2)! &=(1\times\cdots\times n)\times((n+1)\times\cdots\times(2n))\times(2n+1)\times \cdots\times n^2\cr &>(1\times\cdots\times n)\times(1\times\cdots\times n)\times1 \times\cdots\times1\cr &=(n!)^2\ .\cr}$$

Comment. Note that in both solutions we just replace many of the factors by $1$. We can do this because $(n^2)!$ is not just bigger than $(n!)^2$, it is way bigger.

Answer 2

Hint: Note that $$(k^2+2k+1)(k^2+2k)(k^2+2k-1)\cdots (k^2+1)(k!)^2 = [(k^2+2k)(k^2+2k-1)\cdots (k^2+1)][(k^2+2k+1)(k!)^2].$$

Answer 3

You need two factors of $(k+1)$ to turn $(k!)^2$ into $((k+1)!)^2$. Note that $$ (k^2 + 2k + 1) = (k+1)^2 $$ so there are your two factors of $(k+1)$. The rest is just some integer, which is certainly $\ge 1$.