
Asymptotics.
I'm trying to order some numbers....For n very large, I need to order:
2^n, 4^n, 2^(n^2), n!, and n^n
I know 2^n<4^n<n^n, but I'm not sure where the 2^(n^2) and the n! come in...I want to say it goes 2^n<4^n<2^(n^2)<n!<n^n, but I'm not too sure.
Is this correct?

Logarithms will buy you the first three.
$\displaystyle n\cdot log(2)$
$\displaystyle 2n\cdot log(2)$
$\displaystyle n^{2}\cdot log(2)$
That is now rather easy to order.
$\displaystyle 2^{n}\;<\;4^{n}\;<\;2^{n^{2}}$
The last to are easy enough by themselves. One is n factors of n and the other is n factors, but only the largest is as great as n.
$\displaystyle n!\;<\;n^{n}$
Two logs will settle everything but the factorial. You do that.
All we've left is the factorial vs. the $\displaystyle 2^{n^{2}}$
No ideas? Search the brain.
Note: The log is useful because it is monotonic increasing. This is a useful transformation that does not change the order of anything.

Actually, I think I have it. 2^(n^2) is the biggest of all of them since 2^(n^2)=2^(n*n)=(2^n)^n, which is larger than n^n.

$\displaystyle n^{2} \cdot log(2)$
$\displaystyle 2\cdot log(n) + log(log(2))$
vs.
$\displaystyle n \cdot log(n)$
$\displaystyle log(n) + log(log(n))$
Determination:
$\displaystyle log() >> log(log()), $so essentially, we have $\displaystyle 2 \cdot log(n) > log(n)$ and I think you have it. (But I am still a little worried about just how big 'n' is.)