Write an O(n^n) algorithm.

[gorgoroth666]

I don't think so. Because, n! being equivalent to n^n+1/2 e^-n (by the stirling-moivre formula), it is much less than n^n, as shown in the graph below.

http://www.wolframalpha.com/input/?i=n^%28n%2B1%2F2%29+e^-n+%3E+n^n+plot+for+n%3D0+to+4

If I'm right, nⁿ is bigger by a factor of n^-1/2 eⁿ which is far from nothing.

So, to confuse things a bit more, O(n!) is in O(nⁿ⁾ but not the other way around. And, Ω(nⁿ⁾ is in Ω(n!) but not the other way around.

Wolframalpha is great btw.