Upper bound for $n!$

Question

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.

Answer 1

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 $n_0$.