Q:

Prove $\log(x) < n(x)^{1/n}$, for all positive integer values of $n$, and $x > 0$

Given that $$lg(u) < u$$ is always true, how do we use that to prove that $$lg(x) < n(x)^\frac 1n$$
These are the steps that I have taken so far:
$$1: lg(x) < n(x)^\frac 1n$$
$$2: \frac {lg(x)}n < x^\frac 1n$$
$$3: lg(\frac {lg(x)}n) < lg(x)^\frac 1n$$
$$4: lg(\frac {lg(x)}n) < \frac 1n lg(x)$$
$$5: \frac {lg(x)}n < x$$
$$6: lg(x) < nx$$
In which case, since $$lg(u) < u$$ is always true, it thus proves that $$lg(x) < nx$$
I know that I'm probably doing something wrong, especially moving between steps 4 and 5, because the $$\frac 1n$$ shouldn't just disappear like that, but aside from doing this I'm totally stumped. 
Any help in understanding this would be greatly appreciated! Thanks, guys!

A:

Step 4 holds because it is $\log u < u$ with $u={\log x\over n}$, and so the proof ends there, by noticing that steps (1)-(4) are actually reversible.

