Prove that if $t>2n^2$, then $t!>(n^2)^{t-2n^2}$

Question

How to prove that if $t>2n^2$, then $t!>(n^2)^{t-n^2}$, where t is a positive integer and n is an integer greater than 2 ?

I have: $t>2n^2$
$t(t-1)>(2n^2)(2n^2 -1)$
...
$t! > \Pi_{i=0}^{n-1}(2n^2 -i)$

But then how is it linked to the result I would like to prove ?

Answer 1

If $t>2n^2$ then $$\eqalign{t! &=t(t-1)(t-2)\cdots(2n^2)(2n^2-1)\cdots(n^2+1)(n^2)\cdots(2)(1)\cr &\ge t(t-1)(t-2)\cdots(2n^2)(2n^2-1)\cdots(n^2+1)\cr &>(n^2)(n^2)(n^2)\cdots(n^2)(n^2)\cdots(n^2)\cr &=(n^2)^{t-n^2}\ .\cr}$$ The reason for the exponent $t-n^2$ is that we started with $t$ factors and we have thrown out $n^2$ of them.

This is the inequality in your question, and the inequality in your title follows easily.

Comment. If $t>n^2$ this still works: $t>2n^2$ is not necessary. However, if $t>2n^2$ then we can go further: $$t!>(n^2)^{t-n^2}=n^{2t-2n^2}=n^{t+(t-2n^2)}$$ so $$t!>n^t\ ,$$ and I suspect that this may be the ultimate aim of your question.