# Need explanation of answer for finding $f = O(g)$ or $f = Ω(g)$ for $f(n) = [log(n)]^2$, $g(n) = n^{log(n)}$

I have $f(n) = [log(n)]^2$, $g(n) = n^{log(n)}$

And answer $f !=O(g)$, but $f=Ω(g)$, $\forall n>=17$.

With explanation:

$f(n) = e^{nlog(n)}$, $g(n) = e^{log(n)^2}$

$lim_{n->\infty}\frac{log(n)^2}{n}= 0$

I can see on a graphing calculator that that's right, but I'm not sure how you'd find that forall n value with it, or how they got the expressions with the e.

I don't remember doing that sort of transition from calc 3 years ago, and searching for "log e conversions" or "formulas" isn't getting me anything, so I don't even know what to look for. Can some explain how they got this?