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?
 A: 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.
A: The problem is equivalent to determining for which $N$ it holds that $\sqrt[N]{N!} >A$. 
From Stirling approximation it follows that $N! = (N/e)^N \sqrt{  c_N N}$ with $1 < c_N < 2\pi$, and hence $\sqrt[N]{N!} = \frac{N}{e} \sqrt[2N]{ c_N N}$. This means that $ \frac{N}{e} < \sqrt[N]{N!} <  \frac{N}{e} C_k$ for $N \geq k$, where $C_k = \max_{N \geq k}\sqrt[2N]{ c_N N} \leq \sqrt[2k]{ c_k k} $ are constants with $C_k \to 1$ as $k \to \infty$.
Thus, for the optimal value of $N$ you have $\frac{Ae}{C_k} < N < Ae$, provided that you know a priori that $N > k$. In other words, you need to take $N \simeq Ae$, and this approximation gets better as $A$ grows. This gives the answer up to a small multiplicative factor. With a more refined version of Stirling, you might try to get a more precise answer.
Algorithmically speaking, $Ae$ is a good place to start the search for the right $N$. The value  $\sqrt[N]{N!}$ is increasing with respect to $N$, so some form of binary search should work.
