Proving that $f(n) = o(g(n))$ implies $2^{f(n)} = o(2^{g(n)})$ I can't find a counterexample to
$$f(n) = o(g(n)) \text{ implies } 2^{f(n)} = o(2^{g(n)})\,,$$ 
but I do not know a formal way of proving it. Can anyone lead me in the right direction? I asked this exact question yesterday in Stack Overflow by accident (meant to post here) but no one could come up with a valid way of approaching this question.

This is "little-$o$" notation by the way. So a strict upper bound
It should be for monotonically increasing functions
Hint:

We want to prove $2^{f(n)} = o(2^{g(n)})$. This is equivalent to $\lim_{n\to\infty} \dfrac{2^{f(n)}}{2^{g(n)}} = 0$.
We can rewrite the expression inside the limit as $2^{f(n)-g(n)}$, and, exploiting the hypothesis $f(n) = o(g(n))$, we have $\lim_{n\to\infty} f(n)-g(n) = \cdots$