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