threshold of n to satisfy $a^n <n^a$

Question

How to find the minimum of $n$ when we know $a$, to satisfy:

$a^n<n^a$

$a^m>m^a$

for each $m>n$

$n$ and $m$ are natural numbers.

Answer 1

Suppose that $a>1$. One way is to consider the function: $$ f(x)=a^x-x^a $$

The root of this function is obtained as follows: $$ a^x=x^a\implies a^{x/a}=x \implies \frac{-x\ln a}{a}e^{\frac{-x\ln a}{a}}=\frac{-\ln a}{a} $$ Using Lambert W function we have: $$ x^{*}=\frac{-a}{\ln a}W^{-1}(\frac{-\ln a}{a}) $$ The problem is that $W$ is not injective and therefore one cannot properly talk about an inverse function. However we can define $W^{-1}(y)$ as the biggest $x$ such that $W(x)=y$. For $a>1$, you can show that the function is bigger than zero for all $x>x^{*}$.

You can see that $n$, if you are looking for integer $n$, can be chosen as follows: $$ n=\lfloor\frac{-a}{\ln a}W^{-1}(\frac{-\ln a}{a}) \rfloor. $$

Answer 2

Hint: when will you have that $a^n=n^a$? As both $a^n$ and $n^a$ are continuous (for positive $a$ and $n$), there must be a point at which the two are equal.