Proving that $\lim_{n\to\infty} \left(1+\frac{1}{f(n)}\right)^{g(n)} = 1$

We can use that $$ \left(1+\frac{1}{f(n)}\right)^{g(n)} =\left[\left(1+\frac{1}{f(n)}\right)^{f(n)}\right]^{\frac{g(n)}{f(n)}}$$

That depends upon how you defined to grow faster than. But if implies that $\lim_{n\to\infty}\frac{g(n)}{f(n)}=0$,then\begin{align}\lim_{n\to\infty}\left(1+\frac1{f(n)}\right)^{g(n)}&=\lim_{n\to\infty}\left(\left(1+\frac1{f(n)}\right)^{f(n)}\right)^{\frac{g(n)}{f(n)}}\\&=e^0\\&=1.\end{align}

You have $$ \left(1+\frac{1}{f(n)}\right)^{g(n)} = \exp\left(g(n) \ln \left(1+\frac{1}{f(n)}\right) \right) $$ Since $\lim_{n\to\infty} f(n) = \infty$, we have $$ g(n) \ln \left(1+\frac{1}{f(n)}\right) = g(n)\cdot \left(\frac{1}{f(n)} + o\left(\frac{1}{f(n)}\right)\right) = \frac{g(n)}{f(n)} + o\!\left(\frac{g(n)}{f(n)}\right) $$ and by your assumption that $f$ "grows faster than $g$", this converges to $\ell=0$ (the result holds as long as $\lim_{n\to\infty} \frac{g(n)}{f(n)}$ exists, not necessarily $0$).

Then, $$ \lim_{n\to\infty }\left(1+\frac{1}{f(n)}\right)^{g(n)} = e^0 = 1. $$