# How do I prove $f (n) = \omega(2^n)$ if $f(n) = n!$? [closed]

How do I prove $2^n = \Omega(n^2)$?

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## closed as unclear what you're asking by O.L., Macavity, Michael Albanese, T. Bongers, Dominic MichaelisOct 22 '13 at 5:14

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Please write complete sentences. I have no idea what you are trying to say. I don't know what that $w$ is in $f(n)=w(2^n)$. –  Gerry Myerson Oct 23 '11 at 11:11
Do you mean that $n!$ is $\Omega(2^n)$? That at least is true. The rest makes no sense: $n!$ is not $o(a^n)$. It isn’t even $O(a^n)$. It is $\Omega(a^n)$; is that by any chance what you’re trying to show? –  Brian M. Scott Oct 23 '11 at 11:41
This can be generalized to $f(n) = \omega(c^n)$ for all constant $c$. –  Yuval Filmus Oct 23 '11 at 16:25

If you look up the definition of little omega, one definition states that $f(n)=\omega(g(n))$ if \begin{align*} \lim_{n\to\infty}\frac{f(n)}{g(n)}=\infty. \end{align*}
Now, does $\lim_{n\to\infty}n!/2^n$ go to $\infty$?
If you're really lazy, you can always use Stirling's approximation, though it's an overkill unless you're trying to prove things like $n! = \omega(n^{(1-\epsilon)n})$.