The fact that this is almost but not entirely in order of increasing complexity 
https://twitter.com/PPathole/status/1155464941177987072 …
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Try evaluating each one with, say, n=10. That’s enough to sort them into their asymptotic ordering.
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ok, ok. 2^2 = 4 > 2! = 2 Assume a^a > a! for some a. (a+1)! = a! * (a+1) (a+1)^(a+1) = (a+1)^a * (a+1) (a+1)^a > a^a => (a+1)^a > a! => (a+1)^a * (a+1) > a! * (a+1) => (a+1)^(a+1) > (a+1)! => x^x > x! for all x >= a QED.
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Thanks. In this video there is even a simple direct proof: https://www.youtube.com/watch?v=NsO6nh42oPo …pic.twitter.com/H3Y7jhcH5E
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