Finding the solutions of $n! \ge n^a$

Question

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 $\mathbb{N}$?

Answer 1

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.