Question: This is my first ever question here so I hope I present it in a detailed fashion!

We're currently learning about Big O-Notation in college and I happened to come across a training exercise that asked me to prove or disprove the following statement:

$$
n^3 \in 2^{O(n)}
$$

So far I haven't encountered anything like this. Any tips on how to proceed would be really appreciated!

Answer: 
As a set, we can define $2^{O(n)}$ as $\{2^{f(n)} : f(n) \in O(n)\}$.

Alternatively, we can think of these comparisons without writing the set out explicitly, by the definition: $$2^{f(n)} \in 2^{O(n)} \iff f(n) \in O(n).$$
This doesn't seem to directly apply to $n^3$, because it's not in the form $2^{f(n)}$. But we can put it in that form by writing $n^3 = (2^{\log\_2 n})^3 = 2^{3 \log\_2 n}.$

Now the question becomes: is $3 \log\_2 n \in O(n)$? Can you take it from here?
We can interpret the notation as meaning

$$\exists N,c:n>N\implies n^3\le 2^{cn}.$$

This is true: taking the ratio of two successive terms in the progression,

$$\left(\frac{n+1}n\right)^3<2^c$$ is true for $n>1,c=3$

and $n^3$ is bounded above by a geometric progression of common ratio $8$.