why $2^{2n}$ is asymptotically greater than $n!$? My Attempt is that $n!=o(n^n)$..that means it can grow as high as $n^n$ then how it is asymptotically smaller than $2^{2n}$....don't suggest use of calculator!
2026-04-13 16:17:36.1776097056
Asymptotic Notation cormen_doubt
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$2^{2n}$ is not asymptotically greater than $n!$.
If we can use the fact that the series $$ e^4=\sum_{n=0}^\infty\frac{4^n}{n!} $$ converges, then we see that $4^n/n!$ goes to $0$. Hence, $n!$ grows faster than $2^{2n}$.