## Precalculus (10th Edition)

$n!$
We know that $\log_a {x^n}=n\cdot \log_a {x}$ and $\log_a{a}=1$ Hence, $\log_2 {2}\cdot \log_2 {4}\cdot... \log_2 {2^n}\\ =\log_2{2} \cdot \log_2{2^2} \cdot \log_2{2^3} \cdot \log_2{2^4}...\log_2{2^n}\\ =(1\cdot \log_2 {2})\cdot(2\cdot \log_2 {2})\cdot...(n\cdot \log_2 {2})\\ =(1\cdot1) \cdot (2\cdot 1) \cdot (3\cdot 1)...(n\cdot n)\\ =1\cdot2 \cdot 3...\cdot n\\ =n!$