Given that $$x < 2^x$$ is always true, use it to prove that $$x^n < n^n2^x$$
Here are the steps that I've taken so far:
Reduce $$x < 2^x$$ to $$\log(x) < x$$
Then
$$x^n < n^n2^x$$
$$n \log(x) < n \log(n) + x$$
$$\log(x) < \log(n) + \frac {x}{n}$$
And that's where I got stuck.
If it were $$\log(x) < \log(n) + x$$ the proof would be self evident, because since $$\log(x) < x$$ it would be obvious that $(x + \text{anything else})$ would be more than $\log(x)$, but in this case it isn't.
Is there something else that I'm missing? By direct calculation my last step would be true, but how do I relate it to the first statement?
Any help would be greatly appreciated, guys. Thanks a lot!
Let $x\ge 0$. The relation $x^n\lt n^n2^x$ can be rewritten as $\left(\frac{x}{n}\right)^n\lt 2^x$ and then as $\frac{x}{n}\lt 2^{x/n}$. But we are told that $t\lt 2^t$ for all $t$. Put $t=\frac{x}{n}$.
Remark: Let $x=-100$ and $n=2$. Then it is not true that $x^n\lt n^n2^x$.