Order of growth of logarithms, compared to linear I think it is true that any power of a logarithm, no matter how big, will eventually grow slower than a linear function with positive slope.
Is it true that for any exponent $m>0$ (no matter how big we make $m$), the function $f(x)$ $$f(x)=(\ln x)^m$$
will eventually always be less than $g(x) = x$ ?
I am pretty sure this is true because I have tried large values of $m$ and it always eventually slows down. But I don't know how to prove it.

Yes, this is true. This is equivalent to proving that, for any $a > 0$, we have 
$$
\frac{\ln x}{x^a} \xrightarrow[x\to\infty]{}0
$$
(you can see it by setting $a=\frac{1}{m}$ from your question).\; which itself is equivalent to showing
$$
\frac{a\ln x}{x^a} = \frac{\ln x^a}{x^a} \xrightarrow[x\to\infty]{}0
$$
so, at the end of the day, it is sufficient to show ("setting $t = x^a$") that 
$$
\frac{\ln t}{t} \xrightarrow[t\to\infty]{}0.
$$
Now, that you can show in many ways (some would say "L'Hopital's rule"). One of them is to do (yet another) change of variable, $t=e^u$, and show the equivalent statement
$$
\frac{u}{e^u} \xrightarrow[t\to\infty]{}0.
$$
Now, this is obvious given the series definition of the exponential, for instance, as $e^u = 1+u+\frac{u^2}{2}+\sum_{n=3}^\infty \frac{u^n}{n!} > \frac{u^2}{2}$ (the last inequality for $u > 0$), so that $\frac{u}{e^u} < \frac{2}{u}$ for $u>0$.
