Help creating code with running time proportional to N! (factorial)

I'm guessing that n! is something like n^n, but am not sure how to create for loops for that. I know n^2 requires two for loops; one inside the other. Some hints or guides would be greatly appreciated. Thanks.

WootGui wrote:
Thanks for backing up my idea. O(n^2) has two nested loops, that means O(n^n) should have n nested loops, but how is it possible to create n nested loops when n is large. ThanksYou're welcome.
An algorithm does not have to have n-nested loops (which all run upto n) to be in O(n^n). An example of something in O(n^n) (or O(n!)) is solving the famous [travelling salesman problem|http://en.wikipedia.org/wiki/Traveling_salesman_problem] brute force.

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    DIRECTOR> TM_6252 Source Load Summary.
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    DIRECTOR> CMN_1740 Table: [W_PSFT_PAYROLL_F_TMP] (Instance Name: [W_PSFT_PAYROLL_F_TMP])
         Output Rows [0], Affected Rows [0], Applied Rows [0], Rejected Rows [0]
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    DIRECTOR> TM_6020 Session [SDE_PSFT_PayrollFact_Total_Full] completed at [Wed Dec 10 13:41:48 2008]

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