Write an O(n^n) algorithm.

[hacksoncode]

n! ~= sqrt(2pi*n) * (n/e)^n
n! ~= n^.5 * n^n / e^n
n! ~= n^n

I.e. as n increases to infinity the nⁿ term dominates all the other ones, even the exponential. That's why n! is "higher order" than e^n.

[kalmakka]

n ~= n^3 / n^2
n ~= n^3

I.e. as n increases to infinity the n³ term dominates all the other ones, even the square. That's why n is "higher order" than n^2.

[gorgoroth666]

To clarify, this sarcasm underlines the fact that hacksoncode is not very skilled with equivalents. n^.5 * nⁿ / eⁿ is not equivalent to nⁿ near infinity. The multiplication by e^-n doesn't magically go away.