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Let $a\in\mathbb{N}$. is there an upper bound be for the smallest n so that $n!>a$?

It doesn't have to be a good upper bound, just something that works.

Thanks.

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    $\begingroup$ See Stirling's approximation. $\endgroup$ – Lucian Apr 9 '14 at 12:29
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    $\begingroup$ I think $a$ would be an upper bound for such an $n$, at least for $a>2$. As you say, not a good upper bound asymptotically, but certainly an upper bound. $\endgroup$ – Dustan Levenstein Apr 9 '14 at 12:31
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We observe that $$ n!> 2^n \quad (n\geq 4). $$ Indeed, for $n=4$ we have $4!=24>2^4=16$. Suppose that the inequality holds for $n=k$. Then $$ (k+1)!=k!(k+1)>2^k(k+1)>2^k2=2^{k+1}. $$ Choose $n_0\in\mathbb{N}$ smallest such that $2^{n_0}>a$. Then $$ n_0=\left \lfloor{\frac{\ln(a)}{\ln(2)}}\right \rfloor+1. $$

  • If $0\leq a<1$ then the least upper bound for the smallest $n$ such that $n!>a$ is $0$.

  • If $1\leq a<2$ then the least upper bound for the smallest $n$ such that $n!>a$ is $2$.

  • If $2\leq a<6$ then the least upper bound for the smallest $n$ such that $n!>a$ is $3$.

  • If $6\leq a<24$ then the least upper bound for the smallest $n$ such that $n!>a$ is $4$.

  • If $a\geq 24$ then an upper bound for the smallest $n$ such that $n!>a$ is $n_0$.

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