Proving $n\leq3^{n/3}$ for $n\geq0$ via the Well-Ordering Principle [2] I know this question was already asked in here, but it was never marked as answered and all the solutions base themselves on the fact that $3(m-1)^3 < m$, what comes from assuming $3^m < m$ and it's not clear to me.
I tried multiple ways to understand why this assumption was made, but I can't figure it out. My first assumption was that since $m \in S$ it's true that:
$m > 3^{m/3}$
and by consequence:
$m^3 > 3^{m}$
but it still does not prove $3^m < m$. Any help is very appreciated.

The well ordering principle comes into play in trying to find the first $n$ where $n^3 > 3^n$.
We know $1^3 < 3^1$.
But can there be any $n^3 > 3^n$?
If so, there must be a first $n$ where $n^3 > 3^n$. But if $n$ is the first then it must be that  $(n-1)^3 \le 3^{n-1}$.
Now hopefully we will be able to show $(n-1)^3 \le 3^{n-1}$ while $n^3 > 3^n$ can't ever happen which means we can never have a first case where $k^3 \le 3^k$ is not true, which means $k^3 \le 3^k$ always will be true.
So multiplying both sides by $3$ we have$3(n-1)^3 \le 3^n$. But we also have $3^n < n^3$.  So $3(n-1)^3 \le 3^n < n^3$.
And $3(n^3 - 3n^2 + 3n -1) < n^3$.
Well now simplify that and
$2n^3 - 9n^2 + 9n - 3< 0$ so
$2n^3 +9n < 9n^2 + 3$.
$2n + \frac 9n < 9 + \frac 3{n^2}$.
So $2n < 2n + \frac 9n < 9 + \frac 3{n^2} $.
If $n \ge 2$ then $\frac 3{n^2} < 1$ and so we either have $n =1$ or $2n < 9+\frac 3{n^2}< 9+1 = 10$.  In any event we must have $n < 5$ or in other words, $n=1,2,3,$ or $4$.
So we test that $n=1,2,3,4$ and get $1^3 < 3^1; 2^3 =8 < 9 =3^2; 3^3 = 3^3; 4^3=64 < 81 = 3^4$.  so none of those are the first exception.
So the first exception can not exist.
And if there can't be a first value, there can't be any value.
