I am trying to prove the relationship mentioned in the title, only using the definition of $\omega$, but am getting stuck after taking the log of both sides of the inequality:
$$cn^{2} \leq 2^{n}$$
I cannot seem to isolate for n, in terms of c, so that I can show that there exists a value of n, such that for all values of n above that, $n^{2}$ is smaller than $2^{n}$
The limit definition might be easier to work with. In doing so, we seek to show
$$\lim_{n \to \infty} \frac{2^n}{n^2} = \infty$$
We can do this by expanding $2^n$ in a series:
$$2^n = e^{\ln(2) n} = \sum_{k=0}^\infty \frac{\ln(2)^k \cdot n^k}{k!}$$
Then the ratio is given by
$$\sum_{k=0}^\infty \frac{\ln(2)^k \cdot n^{k-2}}{k!}$$
which clearly is infinite in the limit $n \to \infty$. (After all, the sum is
$$\frac{c_0}{n^2} + \frac{c_1}{n} + c_2 + c_3 n + \mathcal O(n^2)$$
for constants $c_k = \ln(2)^k/k!$.)
