$(1-x)^y ≈ e^{-xy}$ Here is an approximation I often see in biology articles but don't really understand:
$$(1-x)^y ≈ e^{-xy}$$
I think this $e^{-xy}$ closely approximates $(1-x)^y$ whenever $x$ is small. Can you help me to understand the conditions for this approximation and why this approximation holds true.

When working with real exponents, it is useful to come back to the definition: 
We have $(1-x)^y=e^{y\ln(1-x)}$ 
but, when $x$ is small, $\ln(1-x)\approx -x$ (since the tangent to the graph $y=\ln(1-x)$ has equation $y=-x$), so for any $y$ and for any $x$ small,
we obtain $y\ln(1-x)\approx -xy$.
Using the continuity of $\exp$, we obtain $(1-x)^y\approx e^{-xy}$.
