What is the mistake in my derivation of the wrong asymptotic relation $n^n = O(n!)$?
Since $n^n = e^{n\log(n)}$, the Taylor series of $e^n$ gives $$ n^n = 1+n\log(n)+O(n^2 \log^2(n)). $$
And $n\log(n) \in O(n!)$ and $ O(n^2 \log^2(n)) \subset O(n!)$, so $n^n = O(n!)$, which is ridiculous.
Could you please point out any mistakes in my calculation? Thank you in advance. Any comment or answer is appreciated.
For $x$ large, the terms in the Taylor series $$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \dots $$ grow larger and larger, so truncating the series will give a wildly wrong estimate (and it will get more wrong the larger $x$ gets).