Prove $\prod_{c < p \leq x} \left(1 - \frac{c}{p} \right) \ll \log ^{-c} x .$ 
For fixed $c>0$, show
  $$\prod_{c < p \leq x} \left(1 - \frac{c}{p} \right) \ll \log ^{-c} x ,$$
  where $p$ is a prime.

I am not sure how to show this result. We know that $$\sum_{p \leq x} \log \left( 1-\frac{1}{p} \right) = -\log \log x - B + O\left (\frac{1}{\log x} \right)$$ for some constant $B>0$, but I am not sure if this helps much as the RHS does not seem to give a hint of the RHS of what we want to prove.

Note that
$$\frac{t}{1+t} < \log (1+t) < t$$
for $t> -1$ and $t \neq 0$. Then
\begin{align*}
c\log \left( 1 - \frac{1}{p} \right) & > \frac{-c/p}{1 - 1/p} \\ & =
\frac{p}{p-1} \left(- \frac{c}{p} \right) \\ & >
  - \frac{c}{p} \\ &
> \log \left(1 - \frac{c}{p} \right)
\end{align*}
for $p >c$. Exponentiate both sides to obtain
$$\left( 1 - \frac{c}{p} \right) < \left( 1 - \frac{1}{p} \right) ^c.$$
So,
\begin{align*}
\prod_{c<p<x} \left(1 - \frac{c}{p} \right) & <
\prod_{c<p<x} \left(1 - \frac{1}{p} \right) ^c \\ & \ll
\prod_{p<x} \\ & \ll
\log ^{-c} x
\end{align*}
