Condition for $N! > A^N$ I am given $A$ , I need to find minimum value of $N$ such that the condition $N! > A^N$ holds.
EXAMPLE : If $A=2$ then minimum $N=4$ and similarly if $A=3$ then minimum $N=7$.
How to solve this problem?

Solve
$$({\frac{n}{e})}^n\sqrt{2\pi n}-A^n = 0$$
with numerical methods, for example the bisection method.
The approximation of n! is good enough even to solve the A=2-case.
The uprounded result is the desired number.
Since
$$n^n > n!$$
for all $n>1$, the desired number must be >A .
So, by induction you can easily prove that the inequality also holds for 
all larger numbers.
Since 
$$(3A)! > A^{3A}$$ 
for any $A\ge1$ (because of $n!>(\frac{n}{3})^n$ , which follows easily
from stirlings formula), 3A is an upper bound for all A.
