Prove by Induction: Weak Inquality involving Product of Factorial, Factorial in the Exponent, Exponents. 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.. 

The inequality is not true. For instance, for $n=3$, we get that $$3! x^{3!} \leq 3 x^3$$ $$6x^6 \leq 3 x^3$$ $$x^3 \leq \frac12$$ The above is true only for $$x \in \left[0,\frac1{2^{1/3}} \right]$$
