A formula for the smallest k such that n^k > n! How to define the smallest exponent k for that n^k > n! ?
Does someone know whether exists a formula for that? 
For example: 
for n = 2, k = 2, since 2^2 > 2!
for n = 3, k = 2, since 3^2 > 3!
for n = 4, k = 3, since 4^3 > 4! 
Thanks in advance

You may solve
$$
n^x>n!
$$ giving
$$
\begin{align}
x \ln n> \ln (n!)
\end{align}
$$ and, for $n=2,3,\ldots$,
$$
x>\frac{\ln (n!)}{\ln n}.
$$ Then
$$
k:=k_n=\left\lceil\frac{\ln (n!)}{\ln n}\right\rceil.
$$ As $n \to \infty$, this may be approximated by using Stirling's formula.
