Finding the solutions of $n! \ge n^a$ Let $a \in \mathbb{N}, a \ge 2$ be a fixed natural number. 
Consider the inequality:
$$n! \ge n^a$$
It can be proven that this inequality is true for sufficiently large values of $n$, but how can we determine its set of solutions in $\mathbb{N}$?

Essentially
you want to solve
$n! = n^a$.
For a first approximation,
you can use
$(n/e)^n < n! < (n/e)^{n+1}
$.
The first gives
$a \ln(n)
> n(\ln(n)-1)
$
so
you want
$a >
n-n/\ln(n)
$.
A first approximation is
$n = a$.
Putting that in
$n
= a+n/\ln(n)
\approx a+a/\ln(a)
=a(1+1/\ln(a))
$.
Try this and see how good it is.
