# $t > 2n^2 \implies t!>n^t$ for $n,t \in \mathbb{N}$

I have come across this in a proof:

If $t>2n^2$ then, $$t!>(n^2)^{t-n^2}=n^tn^{t-2n^2}>n^t$$ Obviously, this is much help to determine the relationship between factorials and exponential, but I fail to see the motivation behind the initial assumption.

I don;t get why $(n^2)t-n^2=n^tn^{t-2n^2}$ – Norbert Jun 7 '12 at 19:49
Sorry, typo, fixed to $(n^2)^{t-n^2}$ – zzzzzzzzzzz Jun 7 '12 at 19:51
Suppose $t > 2n^2$, and for simplicity assume $t$ is even. Then $t/2 > n^2$, so all numbers between $t$ and $t/2$ are bigger than $n^2$. And surely the numbers between $t/2 - 1$ and $1$ are all at least $1$. So
$$t! = \left[t (t-1) (t-2) \ldots \left(\frac{t}{2} + 1\right) \frac{t}{2}\right] \cdot \left(\frac{t}{2} - 1\right)! > \left[n^2 n^2 \cdots n^2\right] \cdot 1 > n^t.$$