Consider the function $z(n) = (1-f(n))^{g(n)}$. For $f(n) = \frac 1n, g(n) = n$ we have that $\lim z(n) = e^{-1}$; more generally, when $f(n) = \frac cn$ for any constant $c$, we have $\lim z(n) = e^{-c}$. In each of these cases we note that the limit is equal to $e^{-f(n)g(n)}$.

Here's my question: Under what conditions (on $f,g$) can we claim that $(1-f(n))^{g(n)} \sim e^{-f(n)g(n)}$? Unless $f(n)g(n)$ is a constant, the RHS also depends on $n$, which is why I'm only asking about asymptotic equality. My original guess was that it is sufficient to have $f(n) \to 0$ and $g(n) \to \infty$ but I haven't been able to get anywhere with a proof.

My context: I have the quantity $( 1 - \frac 1n)^{cn \log n}$ and I'd like desperately for this to be asymptotically equal to $e^{-c\log n} = n^{-c}$.