Relationship between $\omega$ and o

If

$$ne^{\sqrt{\log n}}=o(n^{1+\epsilon})$$

Is true, it means that we need to prove

$$ \lim_{n \to \infty} \dfrac{ne^{\sqrt{\log n}}}{n^{1+\epsilon}} = 0 $$

Now cancelling $n$ in the left expression we have

$$ \lim_{n \to \infty} \dfrac{e^{\sqrt{\log n}}}{n^{\epsilon}} $$

This limit is $\dfrac{\infty}{\infty}$, using L'Hôpital's rule we can derivate and get

$$ \dfrac{d}{dn} \dfrac{e^{\sqrt{\log n}}}{n^{\epsilon}} = \dfrac{-\epsilon e^{\sqrt{\log(n)}}}{n^{\epsilon + 1}} + \dfrac{e^{\sqrt{\log(n)}}}{2n^{\epsilon + 1}\sqrt{\log(n)}} $$

Now remember that if two sequences limits are convergent then the sum or subtraction is also convergent and it converges to the sum or subtraction of the independent limits. We also know by definition that $e^{log{n}} = n$ and from this $e^{\sqrt{\log{n}}} \leq n$ so we can try to evaluate both limits separately and see if we get something that makes sense.

Lets do the first one

$$ \lim_{n \to \infty} \dfrac{\lvert -\epsilon e^{\sqrt{\log(n)}}\rvert}{\lvert n^{\epsilon + 1}\rvert} \leq \lim_{n \to \infty} \dfrac{\epsilon n}{n^{\epsilon + 1}} = \lim_{n \to \infty} \dfrac{\epsilon}{n^{\epsilon}} = 0 $$

And in the same manner you can also attack $\dfrac{e^{\sqrt{\log(n)}}}{2n^{\epsilon + 1}\sqrt{\log(n)}}$ to get the limit is $0$ and thus the subtraction is valid and the original limit is $0$, proving that $$ne^{\sqrt{\log n}}=o(n^{1+\epsilon})$$

$\omega$ is basically the inverse of $o$, so in this case if $ne^{\sqrt{\log n}}=\omega(n\log^c n)$ then $$n\log^c n = o(ne^{\sqrt{\log n}})$$