Show that $ (b_{k})$ is strictly decreasing

Continuing your idea:

$\frac{b_{k+1}}{b_{k}} = (1+ \frac{1}{k+1}) (1 - \frac{1}{(k+1)^2})^{k+1} \stackrel{\text{Bernoulli}}{\le} \left((1+ \frac{1}{(k+1)^2})(1-\frac{1}{(k+1)^2})\right)^{k+1}\stackrel{\text{GM-AM}}{\le} 1^{k+1}=1$

since Use AM-GM inequality we have $$(1+\dfrac{1}{k})^k=(1+\dfrac{1}{k})(1+\dfrac{1}{k})\cdots(1+\dfrac{1}{k})\cdot 1\le\left(\dfrac{k+1+\dfrac{k}{k}}{k+1}\right)^{k+1}=\left(1+\dfrac{1}{k+1}\right)^{k+1}$$