Proving that $f(n) = o(g(n))$ implies $2^{f(n)} = o(2^{g(n)})$
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 $\lim_{n\to\infty} f(n)-g(n) = \cdots$