We use the change-of-base identity $ \log_a{b}=\frac{\log{b}}{\log{a}} $ to find $$\log_2{3} \cdot \log_3{4} \cdot \log_4{5} \cdot \log_5{6} \cdot \log_6{7} \cdot \log_7{8}= \frac{\log3}{\log2} \cdot \frac{\log4}{\log3} \cdot \frac{\log5}{\log4} \cdot \frac{\log6}{\log5} \cdot \frac{\log7}{\log6} \cdot \frac{\log8}{\log7}.$$Simplifying, we get $ \frac{\log8}{\log2}=\log_2{8}=\boxed{3} $.