I proved this by showing that inequality $0\leq n! \leq 2^{2^n}$ always holds using induction.

Is it sufficient to prove $2^{2^n} \in \Omega(n!)$? Do I need to show that $\displaystyle\lim_{n\to\infty}\frac{2^{2^n}}{n!} > 0$ also holds? If yes, then why is it necessary to show this limit (or why it is not sufficient to show the inequality above holds)?

  • $\begingroup$ It's sufficient. Note your inequality shows the limit (if it exists) is $\geq 1$. $\endgroup$ – Wojowu Sep 15 '18 at 16:45
  • $\begingroup$ @Wojowu, thank you for the reply! I can't see the relation, could you please elaborate? $\endgroup$ – Turkhan Badalov Sep 15 '18 at 16:46
  • $\begingroup$ @Wojowu, oh I see, $\frac{2^{2^n}}{n!} \geq 1 \geq 0$ $\endgroup$ – Turkhan Badalov Sep 15 '18 at 16:59

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