# Upper bound for $n!$

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.

• – Lucian Apr 9 '14 at 12:29
• 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. – Dustan Levenstein Apr 9 '14 at 12:31

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 $n_0$.