Good upper bound for $(1-x)^r$

let $a:=\log\left(\frac{1}{1-x}\right)$. Then

$$(1-x)^r=\frac{1}{e^{ra}}=\frac{1}{1+ra+\frac{a^2r^2}{2!}+\frac{a^3r^3}{3!}+\ldots}$$

Now truncate the infinite series in the denominator to your heart's content. For example

$$(1-x)^r\leq \frac{1}{1+ra}$$.

The following inequality may be easier to deal with depending on what you're trying to use it for $$(1-x)^r \leq e^{-xr}.$$

The reason is that $1 + x \leq e^{x}$ for all real $x$. (Note that the $r \geq 0$ assumption is crucial for this inequality to be true.)