# Prove $x^n < n^n 2^x$

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!

• Have you tried to prove it by induction over $n$? I'm not sure but it could work. – Aretino Sep 4 '15 at 5:41
• To expand on André's remark: the LHS becomes arbitrarily large (and positive) also for arbitarily large in absolute value, but negative, $x$ and even $n$. On the contrary, if we fix $n$ the RHS becomes arbitrarily small, and so there's plenty of counterexamples of this kind. – Vincenzo Oliva Sep 4 '15 at 6:41

## 1 Answer

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 $x^n\lt n^n2^x$.