I want to show the below statement.
$n!x^{n!} \leq n\cdot x^{n}$
Where $0 \leq x <1$ for all $n\in \mathbb{N}$
My approach is induction.
The Induction start
This is easy. Set n=1 and the statement $n!x^{n!} \leq n\cdot x^{n}$ is true.
Induction step
Now I assume that $(n-1)!x^{(n-1)!} \leq (n-1)\cdot x^{n-1}$ is true. But I'm stuck here. I have tried to deduce the statement $n!x^{n!} \leq n\cdot x^{n}$ but my attempts seem to fail..